gr-gifted-amateur4-1-32c87219-41b6-4b61-8afa-26755c9bb10f

26

26.1 Singularities 272
26.2 Eddington-Finkelstein coor-
dinates 275
Chapter summary 279
Exercises 279
26.1 Singularities 272 26.2 Eddington-Finkelstein coor- dinates 275 Chapter summary 279 Exercises 279 | 26.1 | Singularities | 272 | | :--- | :--- | ---: | | 26.2 | Eddington-Finkelstein | coor- | | dinates | 275 | | | Chapter summary | 279 | | | Exercises | 279 | |
\curvearrowright This chapter and the next one deal with the technicalities of finding good coordinates suitable to deal with blackhole physics. Those wanting to hole physics. Those wanting to
see more of the physical consesee more of the physical conse-
quences of black holes can skip quences of black holes can skip
to Chapter 28.
1 1 ^(1){ }^{1}1 Although closed trapped surfaces were conceived as part of the discussion of singularities, the existence of a trapped surface doesn't necessarily imply that there is a singularity present. However, when a trapped surface is formed, it does at least suggest the presence of a singularity.
Fig. 26.1 A construction to illustrate closed trapped surfaces, which occur when the area of both S 1 S 1 S_(1)S_{1}S1 and S 2 S 2 S_(2)S_{2}S2 are smaller than that of S S SSS.

Black-hole singularities

If the doors of perception were cleansed every thing would appear to man as it is, Infinite. For man has closed himself up, till he sees all things thro' narrow chinks of his cavern. William Blake (1757-1827)
In 1965, Roger Penrose described the geometry of the inner region of a black hole's event horizon using the concept of a closed trapped surface. The argument goes like this. The 2 -sphere S S SSS encloses a large spherical distribution of mass. Let a point P P P\mathcal{P}P on S S SSS emit light pulses. At some later time the pulse will form a sphere T T TTT around P P P\mathcal{P}P. If all points on S S SSS emit pulses then the incoming and outgoing wavefronts will have envelopes forming two 2 -spheres, S 1 S 1 S_(1)S_{1}S1 and S 2 S 2 S_(2)S_{2}S2, respectively, as shown in Fig. 26.1. Normally the area of the sphere S 1 S 1 S_(1)S_{1}S1 (formed from all of the incoming parts) must be less than that of S S SSS, while S 2 S 2 S_(2)S_{2}S2, formed from the outgoing parts, will be larger. However, when the amount of mass enclosed by S S SSS is sufficiently large, the areas of S 1 S 1 S_(1)S_{1}S1 and S 2 S 2 S_(2)S_{2}S2 will both be less than S S SSS. The surface S S SSS is then known as a trapped surface and all outward-going light rays will converge towards the centre of the mass distribution. Clearly something significant has caused the trapped surface: here it is a singularity at the centre of a black hole.
This argument illustrates some of the strange behaviour that can occur near a physical singularity in spacetime. 1 1 ^(1){ }^{1}1 In this chapter, we look in more detail at the singularities occurring in the physics of black holes. As we shall see, the identification of genuine singularities is rather difficult, hence the need for physical arguments like the one above.

26.1 Singularities

One difficulty in dealing with the metric field of general relativity is the occurrence of singularities. A singularity is, roughly speaking, a point where a mathematical object is undefined, or fails to be well-behaved in some way. An example might be curvature blowing up or some other pathological behaviour in the metric line element. Near a true, physical singularity in the curvature of the metric, it is impossible to find an area small enough that space looks locally flat. (If in doubt consider the surface of a cone close to its apex). We therefore lose the smoothness property of spacetime, putting our description in terms of fields at risk. Physically, such a singularity is a point where our laws of physics break down.
Some singularities are more harmful that others. The less harmful variety, known as a coordinate singularity simply reflect our choice of coordinates. These are still a problem since they affect our ability to work with the geometry near the singular point, but they can effectively be eliminated. An instance of a coordinate singularity is given in the next example. 2 2 ^(2){ }^{2}2

Example 26.1

Two-dimensional cylindrical coordinates have line element
(26.1) d s 2 = d r 2 + r 2 d θ 2 (26.1) d s 2 = d r 2 + r 2 d θ 2 {:(26.1)ds^(2)=dr^(2)+r^(2)dtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2} \tag{26.1} \end{equation*}(26.1)ds2=dr2+r2 dθ2
These coordinates become singular at the point r = 0 r = 0 r=0r=0r=0. Here, changes in θ θ theta\thetaθ become meaningless, all values of θ θ theta\thetaθ apparently labelling a single point at r = r = r=r=r= 0 . Put more vividly: there is a breakdown in the meaning of our coordinates. However, expressed in two-dimensional Cartesian coordinates, we have the
usual line element d s 2 = d x 2 + d y 2 d s 2 = d x 2 + d y 2 ds^(2)=dx^(2)+dy^(2)\mathrm{d} s^{2}=\mathrm{d} x^{2}+\mathrm{d} y^{2}ds2=dx2+dy2. The Cartesian coordinates are well behaved around the origin, showing that the problem we identified from the cylindrical coordinate description does not reflect a problem with the spacetime. 3 3 ^(3){ }^{3}3
The Schwarzschild metric 4 4 ^(4){ }^{4}4 with which we are concerned in this chapter is singular in two places: at r = 0 r = 0 r=0r=0r=0 and at r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M. The singularity at r = 0 r = 0 r=0r=0r=0 is a genuine physical singularity reflecting an infinite curvature of spacetime. (In this case, it turns this can be demonstrated by evaluating a scalar 5 R μ ν α β R μ ν α β 5 R μ ν α β R μ ν α β ^(5)R_(mu nu alpha beta)R^(mu nu alpha beta){ }^{5} R_{\mu \nu \alpha \beta} R^{\mu \nu \alpha \beta}5RμναβRμναβ built from the components of the Riemann tensor, which diverges there.) In contrast, the singularity at the event horizon at r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M is a coordinate singularity, reflecting a breakdown in the Schwarzschild-coordinate description of the spacetime. The point of this section is to examine how to remove the apparent singularity so that we can describe the physical behaviour close to the event horizon at r S r S r_(S)r_{\mathrm{S}}rS.
To understand coordinate singularities it's useful to have access to another conceptual tool. Some metric spaces are geodesically complete. This means that a geodesic, described by a function x μ ( λ ) x μ ( λ ) x^(mu)(lambda)x^{\mu}(\lambda)xμ(λ), can be continued out to arbitrary points in space by varying the affine parameter λ λ lambda\lambdaλ. In contrast, some spaces are geodesically incomplete: an attempt to extend the geodesics hits a badly defined point. This is shown in Fig. 26.2. In general relativity, this idea allows a more robust definition of a singularity than the rough one offered so far: a singularity occurs in a spacetime if there are incomplete geodesics.
Example 26.2
Consider a spacetime with line element
(26.3) d s 2 = 1 t 4 d t 2 + d x 2 (26.3) d s 2 = 1 t 4 d t 2 + d x 2 {:(26.3)ds^(2)=-(1)/(t^(4))dt^(2)+dx^(2):}\begin{equation*} \mathrm{d} s^{2}=-\frac{1}{t^{4}} \mathrm{~d} t^{2}+\mathrm{d} x^{2} \tag{26.3} \end{equation*}(26.3)ds2=1t4 dt2+dx2
which appears to be badly behaved at t = 0 t = 0 t=0t=0t=0 and t = t = t=oot=\inftyt=. First, there appears to be a singularity at t = 0 t = 0 t=0t=0t=0. However, if we make the coordinate transformation y = 1 t y = 1 t y=(1)/(t)y=\frac{1}{t}y=1t, we find that
(26.4) d s 2 = d y 2 + d x 2 (26.4) d s 2 = d y 2 + d x 2 {:(26.4)ds^(2)=-dy^(2)+dx^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} y^{2}+\mathrm{d} x^{2} \tag{26.4} \end{equation*}(26.4)ds2=dy2+dx2
2 2 ^(2){ }^{2}2 We follow the approach of Wald in the examples in this chapter, which can be consulted for further details.
3 3 ^(3){ }^{3}3 In many cases, coordinate singularities mean that several sets of overlapping coordinates are needed to cover a region of spacetime.
4 4 ^(4){ }^{4}4 Reminder: The Schwarzschild metric line element is given by
d s 2 = ( 1 2 M r ) d t 2 (26.2) + ( 1 2 M r ) 1 d r 2 + r 2 d Ω 2 d s 2 = 1 2 M r d t 2 (26.2) + 1 2 M r 1 d r 2 + r 2 d Ω 2 {:[ds^(2)=-(1-(2M)/(r))dt^(2)],[(26.2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)dOmega^(2)]:}\begin{align*} \mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2} \\ & +\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2} \mathrm{~d} \Omega^{2} \tag{26.2} \end{align*}ds2=(12Mr)dt2(26.2)+(12Mr)1 dr2+r2 dΩ2
where d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ2=dθ2+sin2θ dϕ2
The components of the Riemann tensor in the orthonormal frame are
R τ ^ r ^ γ ^ r ^ = 2 M r 3 , R θ ^ ϕ ^ θ ^ ϕ ^ = + 2 M r 3 , R γ ^ θ ^ γ ^ θ ^ = R τ ^ ϕ ^ γ ^ ϕ ^ ^ = + M r 3 , R r ^ θ ^ r ^ θ ^ = R r ^ ϕ ^ r ^ ϕ ^ = M r 3 . R τ ^ r ^ γ ^ r ^ = 2 M r 3 , R θ ^ ϕ ^ θ ^ ϕ ^ = + 2 M r 3 , R γ ^ θ ^ γ ^ θ ^ = R τ ^ ϕ ^ γ ^ ϕ ^ ^ = + M r 3 , R r ^ θ ^ r ^ θ ^ = R r ^ ϕ ^ r ^ ϕ ^ = M r 3 . {:[R_( hat(tau) hat(r) hat(gamma) hat(r))=-(2M)/(r^(3))","],[R_( hat(theta) hat(phi) hat(theta) hat(phi))=+(2M)/(r^(3))","],[R_( hat(gamma) hat(theta) hat(gamma) hat(theta))=R_( hat(tau) hat(phi) hat(gamma) hat(phi) hat())=+(M)/(r^(3))","],[R_( hat(r) hat(theta) hat(r) hat(theta))=R_( hat(r) hat(phi) hat(r) hat(phi))=-(M)/(r^(3)).]:}\begin{gathered} R_{\hat{\tau} \hat{r} \hat{\gamma} \hat{r}}=-\frac{2 M}{r^{3}}, \\ R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=+\frac{2 M}{r^{3}}, \\ R_{\hat{\gamma} \hat{\theta} \hat{\gamma} \hat{\theta}}=R_{\hat{\tau} \hat{\phi} \hat{\gamma} \hat{\phi} \hat{}}=+\frac{M}{r^{3}}, \\ R_{\hat{r} \hat{\theta} \hat{r} \hat{\theta}}=R_{\hat{r} \hat{\phi} \hat{r} \hat{\phi}}=-\frac{M}{r^{3}} . \end{gathered}Rτ^r^γ^r^=2Mr3,Rθ^ϕ^θ^ϕ^=+2Mr3,Rγ^θ^γ^θ^=Rτ^ϕ^γ^ϕ^^=+Mr3,Rr^θ^r^θ^=Rr^ϕ^r^ϕ^=Mr3.
Notice that the curvature has no unusual behaviour at r = 2 M r = 2 M r=2Mr=2 Mr=2M, but that the components do blow up at r = 0 r = 0 r=0r=0r=0.
5 5 ^(5){ }^{5}5 Why? A scalar is the same in any coordinate system, so if a scalar diverges it suggests there is really a singularity in the physics.
Fig. 26.2 (a) A geodesically complete space where all geodesics can be extended to infinity. (b) A geodesically incomplete space. The hatched area represents a singularity, through which the geodesics can't be extended.
6 6 ^(6){ }^{6}6 We can write y ( λ ) = m x ( λ ) + c y ( λ ) = m x ( λ ) + c y(lambda)=mx(lambda)+cy(\lambda)=m x(\lambda)+cy(λ)=mx(λ)+c and use the parametrization x = λ x = λ x=lambdax=\lambdax=λ so t = t = t=t=t= ( m λ + c ) 1 ( m λ + c ) 1 (m lambda+c)^(-1)(m \lambda+c)^{-1}(mλ+c)1. Make λ λ lambda\lambdaλ arbitrarily large and t 0 t 0 t rarr0t \rightarrow 0t0. The geodesics are complete around here: we can make λ λ lambda\lambdaλ as large as we like and we still return a valid geodesic. One the other hand, as we send λ λ lambda\lambdaλ towards c / m c / m -c//m-c / mc/m we have a problem. Do this from above ( λ = c / m + ε ( λ = c / m + ε (lambda=-c//m+epsi(\lambda=-c / m+\varepsilon(λ=c/m+ε, for ε 0 + ε 0 + epsi rarr0^(+)\varepsilon \rightarrow 0^{+}ε0+) and t t t rarr oot \rightarrow \inftyt; do it from below and t t t rarr-oot \rightarrow-\inftyt. The geodesics are therefore not complete when t t ttt becomes very large. They must stop at positive infinity, preventing us setting λ < c / m λ < c / m lambda < -c//m\lambda<-c / mλ<c/m.
Fig. 26.3 The part of Minkowski space covered by the coordinate sysspace covered by the coordinate sys-
tem in eqn 26.3 . To cover the whole tem in eqn 26.3. To cover the whole
of Minkowski spacetime we need to add on the part below y = 0 y = 0 y=0y=0y=0.
7 7 ^(7){ }^{7}7 We already did this in setting up lightcone coordinates in Minkowski space. These coordinates rely on the definitions
u = t x v = t + x u = t x v = t + x {:[u=t-x],[v=t+x]:}\begin{aligned} u & =t-x \\ v & =t+x \end{aligned}u=txv=t+x
The line element for Minkowski spacetime can be written as
d s 2 = d u d v d s 2 = d u d v ds^(2)=-dudv\mathrm{d} s^{2}=-\mathrm{d} u \mathrm{~d} vds2=du dv
Notice that this is not a diagonal metric. Null geodesics are defined by d s 2 = d s 2 = ds^(2)=\mathrm{d} s^{2}=ds2= 0 . Outgoing null geodesics, for which x x xxx increases as t t ttt increases, have u = u = u=u=u= const. and d u = 0 d u = 0 du=0\mathrm{d} u=0du=0; incoming null and d u = 0 d u = 0 du=0\mathrm{d} u=0du=0; incoming null geodesics
have v = v = v=v=v= const, and d v = 0 d v = 0 dv=0\mathrm{d} v=0dv=0. The null have v = v = v=v=v= const. and d v = 0 d v = 0 dv=0\mathrm{d} v=0dv=0. The null geodesics can therefore be used to make 8 8 _(8){ }_{8}8 g grid covering flat space.
8 8 ^(8){ }^{8}8 See Chapter 5 for a picture of the light-cone structure for this metric.
that is, the metric in eqn 26.3 is Minkowski space in disguise. The apparent singularity at t = 0 t = 0 t=0t=0t=0 corresponds to the point y y y rarr ooy \rightarrow \inftyy in the Minkowski space. The cause of the problem was that all points at infinity in the Minkowski space were being mapped to t = 0 t = 0 t=0t=0t=0 in the original coordinate system of eqn 26.3 , and were therefore not given unique labels.
We can also carry out an analysis using geodesics, which are straight lines in this flat space. 6 6 ^(6){ }^{6}6 The metric in eqn 26.3 is geodesically complete as y y y rarr ooy \rightarrow \inftyy, or as t 0 t 0 t rarr0t \rightarrow 0t0, confirming that t = 0 t = 0 t=0t=0t=0 is merely a coordinate singularity. However, we are not out of the woods since, in the original coordinates, the space is not geodesically complete for large t t ttt. Specifically, because of the form of the metric, the geodesics stop as t t t rarr oot \rightarrow \inftyt. In Minkowski coordinates, t t t rarr oot \rightarrow \inftyt is equivalent to the point y = 0 y = 0 y=0y=0y=0, implying that those original coordinates cannot access the parts of the space corresponding to negative y y yyy.
The cause of this singularity is that the metric in eqn 26.3 does not cover all of Minkowski space; only that portion with y > 0 y > 0 y > 0y>0y>0, since y 0 y 0 y rarr0y \rightarrow 0y0 for t t t rarr oot \rightarrow \inftyt as shown in Fig. 26.3. However, we can extend the spacetime beyond t = t = t=oot=\inftyt=. This is achieved by considering the transformed spacetime in eqn 26.4 and adding the portion y 0 y 0 y <= 0y \leq 0y0 to the spacetime. We see that we were misled by the original coordinate description of this well-behaved spacetime. There is nothing singular about it beyond the coordinate singularity.
The previous example gives us a clue of how to identify, interpret and, where possible, eliminate coordinate singularities. In many applications, such as the black holes examined in this chapter, coordinate singularities are transformed away with an inspired choice of coordinates. Although this often looks like magic, it is well motivated by a sensible method. In the next example, we show how a tricky spacetime with an apparent singularity can be understood with a change of coordinates that aims to use null geodesics as a coordinate system. We will sketch out a generalizable strategy to remove the singular-looking parts of a spacetime. The key is to examine the incoming and outgoing null geodesics and use these as a grid. 7 7 ^(7){ }^{7}7

Example 26.3

Consider Rindler spacetime 8 8 ^(8){ }^{8}8, which has line element
(26.5) d s 2 = x 2 d t 2 + d x 2 (26.5) d s 2 = x 2 d t 2 + d x 2 {:(26.5)ds^(2)=-x^(2)dt^(2)+dx^(2):}\begin{equation*} \mathrm{d} s^{2}=-x^{2} \mathrm{~d} t^{2}+\mathrm{d} x^{2} \tag{26.5} \end{equation*}(26.5)ds2=x2 dt2+dx2
This appears to have a singularity at x = 0 x = 0 x=0x=0x=0. At this point, the determinant of the matrix representing the metric g ( = x 2 ) g = x 2 g(=-x^(2))g\left(=-x^{2}\right)g(=x2) vanishes, making the inverse components, g μ ν g μ ν g^(mu nu)g^{\mu \nu}gμν, singular. In two dimensions, null geodesics divide up into incoming and outgoing. Within each class they do not cross, so make a natural grid that can be used as a coordinate system. The null geodesics in Rindler spacetime may be found by considering d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0, or
(26.6) ( d t d x ) 2 = 1 x 2 (26.6) d t d x 2 = 1 x 2 {:(26.6)((dt)/((d)x))^(2)=(1)/(x^(2)):}\begin{equation*} \left(\frac{\mathrm{d} t}{\mathrm{~d} x}\right)^{2}=\frac{1}{x^{2}} \tag{26.6} \end{equation*}(26.6)(dt dx)2=1x2
so that, along the geodesics,
(26.7) t = ± ln x + const. (26.7) t = ± ln x +  const.  {:(26.7)t=+-ln x+" const. ":}\begin{equation*} t= \pm \ln x+\text { const. } \tag{26.7} \end{equation*}(26.7)t=±lnx+ const. 
where c c ccc is a constant. The plus sign gives the outgoing geodesics, the minus sign the incoming ones. The equations for the light cones, by analogy with the version for Minkowski space, allow us to define a set of null coordinates ( u , v ) ( u , v ) (u,v)(u, v)(u,v)
u = t ln x (26.8) v = t + ln x u = t ln x (26.8) v = t + ln x {:[u=t-ln x],[(26.8)v=t+ln x]:}\begin{align*} & u=t-\ln x \\ & v=t+\ln x \tag{26.8} \end{align*}u=tlnx(26.8)v=t+lnx
Using the null coordinates, we write the metric as
(26.9) d s 2 = e v u d u d v (26.9) d s 2 = e v u d u d v {:(26.9)ds^(2)=-e^(v-u)dudv:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{v-u} \mathrm{~d} u \mathrm{~d} v \tag{26.9} \end{equation*}(26.9)ds2=evu du dv
This looks rather like the Minkowski metric expressed in light-cone coordinates d s 2 = d s 2 = ds^(2)=\mathrm{d} s^{2}=ds2= d u d v d u d v -dudv-\mathrm{d} u \mathrm{~d} vdu dv (albeit with a prefactor). However, it's notable that the coordinates u u uuu and v v vvv only extend over the regions of x > 0 x > 0 x > 0x>0x>0, so we need to re-parametrize the spacetime in order to make it geodesically complete by introducing new variables U ( u ) U ( u ) U(u)U(u)U(u) and V ( v ) V ( v ) V(v)V(v)V(v). Given the resemblance to Minkowski space, it is possible to guess a set of coordinates that will do this. The answer is to transform u u uuu and v v vvv into
(26.10) U = e u , V = e v (26.10) U = e u , V = e v {:(26.10)U=-e^(-u)","quad V=e^(v):}\begin{equation*} U=-\mathrm{e}^{-u}, \quad V=\mathrm{e}^{v} \tag{26.10} \end{equation*}(26.10)U=eu,V=ev
The coordinates U U UUU and V V VVV are allowed to take all values from U , V U , V -oo <= U,V <= oo-\infty \leq U, V \leq \inftyU,V. The metric line element then becomes
(26.11) d s 2 = d U d V (26.11) d s 2 = d U d V {:(26.11)ds^(2)=-dUdV:}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} U \mathrm{~d} V \tag{26.11} \end{equation*}(26.11)ds2=dU dV
which we recognize as the interval for Minkowski space expressed in light-cone coordinates. This space is geodesically complete and free from singularities. To confirm this, we could make a final transformation
(26.12) T = ( U + V ) 2 , X = ( V U ) 2 (26.12) T = ( U + V ) 2 , X = ( V U ) 2 {:(26.12)T=((U+V))/(2)","quad X=((V-U))/(2):}\begin{equation*} T=\frac{(U+V)}{2}, \quad X=\frac{(V-U)}{2} \tag{26.12} \end{equation*}(26.12)T=(U+V)2,X=(VU)2
and conclude that 9 9 ^(9){ }^{9}9
(26.14) d s 2 = d T 2 + d X 2 (26.14) d s 2 = d T 2 + d X 2 {:(26.14)ds^(2)=-dT^(2)+dX^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2} \tag{26.14} \end{equation*}(26.14)ds2=dT2+dX2
We were dealing with flat space all along!
So what was the reason for the coordinate singularity in Rinder spacetime? Rinder spacetime covers only a wedge of Minkowski space: that part with X > | T | X > | T | X > |T|X>|T|X>|T|, as shown in Fig. 26.4. The transformation to U U UUU and V V VVV allowed us to break through the coordinate barrier and extend our coordinates over all of the spacetime.

26.2 Eddington-Finkelstein coordinates

After taming Rindler space, we need to do the same for the space described by the Schwarzschild metric. Our aim is to find a new set of coordinates that shows that the singularity at r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M can be eliminated, just as we eliminated coordinate singularities in Rindler space. Taking tentative steps, we start with a warm-up exercise.

Example 26.4

Consider the following two-dimensional metric line element (for a sort of babySchwarzschild metric)
(26.15) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 (26.15) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 {:(26.15)ds^(2)=-(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2} \tag{26.15} \end{equation*}(26.15)ds2=(12Mr)dt2+(12Mr)1 dr2
and repeat the argument from Rindler space. The null geodesics, found from setting d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0, are written as
(26.16) ( d t d r ) 2 = ( 1 1 2 M r ) 2 (26.16) d t d r 2 = 1 1 2 M r 2 {:(26.16)((dt)/((d)r))^(2)=((1)/(1-(2M)/(r)))^(2):}\begin{equation*} \left(\frac{\mathrm{d} t}{\mathrm{~d} r}\right)^{2}=\left(\frac{1}{1-\frac{2 M}{r}}\right)^{2} \tag{26.16} \end{equation*}(26.16)(dt dr)2=(112Mr)2
Integrating, we find that the radial null geodesics satisfy
(26.17) t = ± r (26.17) t = ± r {:(26.17)t=+-r^(**):}\begin{equation*} t= \pm r^{*} \tag{26.17} \end{equation*}(26.17)t=±r
9 9 ^(9){ }^{9}9 The original coordinates in terms of the final coordinates are
x = ( X 2 T 2 ) 1 2 (26.13) t = tanh 1 ( T X ) . x = X 2 T 2 1 2 (26.13) t = tanh 1 T X . {:[x=(X^(2)-T^(2))^((1)/(2))],[(26.13)t=tanh^(-1)((T)/(X)).]:}\begin{align*} x & =\left(X^{2}-T^{2}\right)^{\frac{1}{2}} \\ t & =\tanh ^{-1}\left(\frac{T}{X}\right) . \tag{26.13} \end{align*}x=(X2T2)12(26.13)t=tanh1(TX).
It would have been quite mysterious to have picked these coordinates out of nowhere.
Fig. 26.4 The part of Minkowski space ( T , X T , X T,XT, XT,X ) covered by the Rindler metric coordinates.
10 10 ^(10){ }^{10}10 Recall from Chapter 25 that
r = r + 2 M ln | r 2 M 1 | + const. r = r + 2 M ln r 2 M 1 +  const.  r^(**)=r+2M ln|(r)/(2M)-1|+" const. "r^{*}=r+2 M \ln \left|\frac{r}{2 M}-1\right|+\text { const. }r=r+2Mln|r2M1|+ const. 
We will set the constant to zero here.
11 11 ^(11){ }^{11}11 David Finkelstein (1929-2016). In addition to his work on quantum gravity, Finkelstein also worked on ball lightning and provided a detailed analysis of Albrecht Dürer's Melencolia I. He was also responsible for naming the sine-Gordon model: a joke, based on the Klein-Gordon model, that he later regretted.
12 12 ^(12){ }^{12}12 As in Schwarzschild coordinates we have the conserved quantity L ~ L ~ tilde(L)\tilde{L}L~. Notice also that there's no dependence on V V VVV in the metric and so in place of a conserved quantity E ~ = u t E ~ = u t tilde(E)=-u_(t)\tilde{E}=-u_{t}E~=ut, we have the conserved quantity E ~ = u V E ~ = u V tilde(E)=-u_(V)\tilde{E}=-u_{V}E~=uV.
Fig. 26.5 Light cones in incoming Eddington-Finkelstein coordinates Here we set t ¯ = V r t ¯ = V r bar(t)=V-r\bar{t}=V-rt¯=Vr and plot the t ¯ r t ¯ r bar(t)-r\bar{t}-rt¯r plane.
where r r r^(**)r^{*}r is the tortoise coordinate. 10 10 ^(10){ }^{10}10 Define null coordinates by
U = t r (26.18) V = t + r U = t r (26.18) V = t + r {:[U=t-r^(**)],[(26.18)V=t+r^(**)]:}\begin{align*} U & =t-r^{*} \\ V & =t+r^{*} \tag{26.18} \end{align*}U=tr(26.18)V=t+r
Outgoing, null geodesics are given by U = U = U=U=U= const. Incoming, null geodesics are given by V = V = V=V=V= const. In these coordinates, the metric becomes
(26.19) d s 2 = ( 1 2 M r ) d U d V (26.19) d s 2 = 1 2 M r d U d V {:(26.19)ds^(2)=-(1-(2M)/(r))dUdV:}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} U \mathrm{~d} V \tag{26.19} \end{equation*}(26.19)ds2=(12Mr)dU dV
where r r rrr is defined implicitly by
(26.20) r + 2 M ln | r 2 M 1 | = r = ( V U ) 2 (26.20) r + 2 M ln r 2 M 1 = r = ( V U ) 2 {:(26.20)r+2M ln|(r)/(2M)-1|=r^(**)=((V-U))/(2):}\begin{equation*} r+2 M \ln \left|\frac{r}{2 M}-1\right|=r^{*}=\frac{(V-U)}{2} \tag{26.20} \end{equation*}(26.20)r+2Mln|r2M1|=r=(VU)2
Equation 26.19 has the form of a function multiplied by the Minkowski light-cone metric, just as we had in the last example. Although we could push through to complete the process, we pause at this point to examine a useful set of halfway-house coordinates.
Inspired by the argument above, one set of coordinates that are often used in the study of black holes are incoming Eddington-Finkelstein coordinates. 11 11 ^(11){ }^{11}11 These coordinates are particularly useful for describing things that fall into back holes. They are a hybrid set of coordinates that use r r rrr and (the incoming null coordinate) V V VVV in place of r r rrr and t t ttt. That is, we substitute for the time coordinate t t ttt in the Schwarzschild metric with
(26.21) t = V r (26.21) t = V r {:(26.21)t=V-r^(**):}\begin{equation*} t=V-r^{*} \tag{26.21} \end{equation*}(26.21)t=Vr
The Schwarzschild metric line element becomes
(26.22) d s 2 = ( 1 2 M r ) d V 2 + 2 d V d r + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (26.22) d s 2 = 1 2 M r d V 2 + 2 d V d r + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(26.22)ds^(2)=-(1-(2M)/(r))dV^(2)+2dVdr+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} V^{2}+2 \mathrm{~d} V \mathrm{~d} r+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{26.22} \end{equation*}(26.22)ds2=(12Mr)dV2+2 dV dr+r2( dθ2+sin2θ dϕ2)
This metric is not diagonal: it has the cross term 2 d V d r 2 d V d r 2dVdr2 \mathrm{~d} V \mathrm{~d} r2 dV dr. There is no component d r 2 d r 2 dr^(2)\mathrm{d} r^{2}dr2, but we shouldn't forget the presence of r r rrr in the calculations. 12 12 ^(12){ }^{12}12

Example 26.5

What do the light cones look like using these coordinates? To see them, set d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0 and find
(26.23) ( 1 2 M r ) d V 2 + 2 d V d r = 0 (26.23) 1 2 M r d V 2 + 2 d V d r = 0 {:(26.23)-(1-(2M)/(r))dV^(2)+2dVdr=0:}\begin{equation*} -\left(1-\frac{2 M}{r}\right) \mathrm{d} V^{2}+2 \mathrm{~d} V \mathrm{~d} r=0 \tag{26.23} \end{equation*}(26.23)(12Mr)dV2+2 dV dr=0
Incoming rays have V = V = V=V=V= const (as we should have expected). The other solution is
(26.24) ( 1 2 M r ) d V + 2 d r = 0 (26.24) 1 2 M r d V + 2 d r = 0 {:(26.24)-(1-(2M)/(r))dV+2dr=0:}\begin{equation*} -\left(1-\frac{2 M}{r}\right) \mathrm{d} V+2 \mathrm{~d} r=0 \tag{26.24} \end{equation*}(26.24)(12Mr)dV+2 dr=0
which we integrate to find that outgoing rays follow
(26.25) V 2 ( r + 2 M ln | r 2 M 1 | ) = const. (26.25) V 2 r + 2 M ln r 2 M 1 =  const.  {:(26.25)V-2(r+2M ln|(r)/(2M)-1|)=" const. ":}\begin{equation*} V-2\left(r+2 M \ln \left|\frac{r}{2 M}-1\right|\right)=\text { const. } \tag{26.25} \end{equation*}(26.25)V2(r+2Mln|r2M1|)= const. 
The Eddington-Finkelstein coordinates therefore use one incoming null coordinate and one tortoise-like coordinate to form a grid for the light cones. Incoming Eddington-Finkelstein coordinates have the advantage that incoming light rays are simply continuous straight lines that head towards the singularity at r = 0 r = 0 r=0r=0r=0, as shown in Fig. 26.5. They show no unusual behaviour passing through r S r S r_(S)r_{\mathrm{S}}rS. The price we pay is that outgoing rays still look singular at r S r S r_(S)r_{\mathrm{S}}rS. The 'outgoing' rays reach infinity if they start outside the horizon, but fall into the r = 0 r = 0 r=0r=0r=0 singularity if they start inside the horizon. Notice from eqn 26.23 that if r = 2 M r = 2 M r=2Mr=2 Mr=2M then we have V = r = V = r = V=r=V=r=V=r= const. The light rays are then trapped at the horizon, neither falling into the hole, nor emerging from it.
Having a well-behaved incoming coordinate V V VVV allows us to continuously describe things falling through the horizon at r S r S r_(S)r_{\mathrm{S}}rS.

Example 26.6

Consider an astronaut standing on the surface of a collapsing star. This is shown in incoming Eddington-Finkelstein coordinates in Fig. 26.6. The surface of the star shrinks radially and, because of the choice of coordinates, looks intuitively as expected. If the astronaut sends out regular light pulses, these follow the lines of constant V 2 r V 2 r V-2r^(**)V-2 r^{*}V2r as shown. We see that the interval between light pulses being received at a large distance from the hole increases. Once the astronaut has fallen through r S r S r_(S)r_{\mathrm{S}}rS the light pulses do not emerge from the hole.
We can also define outgoing Eddington-Finkelstein coordinates, by using r r rrr and U U UUU in place of r r rrr and t t ttt (i.e. we set t = U + r t = U + r t=U+r^(**)t=U+r^{*}t=U+r ). The Schwarzschild metric in these coordinates becomes
(26.26) d s 2 = ( 1 2 M r ) d U 2 + 2 d U d r + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (26.26) d s 2 = 1 2 M r d U 2 + 2 d U d r + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(26.26)ds^(2)=-(1-(2M)/(r))dU^(2)+2dUdr+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} U^{2}+2 \mathrm{~d} U \mathrm{~d} r+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{26.26} \end{equation*}(26.26)ds2=(12Mr)dU2+2 dU dr+r2( dθ2+sin2θ dϕ2)
By evaluating the light cones, we see that one solution to d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0 has U = U = U=U=U= const and the other has
(26.27) d U d r = 2 1 2 M r (26.27) d U d r = 2 1 2 M r {:(26.27)(dU)/((d)r)=-(2)/(1-(2M)/(r)):}\begin{equation*} \frac{\mathrm{d} U}{\mathrm{~d} r}=-\frac{2}{1-\frac{2 M}{r}} \tag{26.27} \end{equation*}(26.27)dU dr=212Mr
Comparing with the incoming version, we see that (as is easily guessed) outgoing Eddington-Finkelstein coordinates have continuous null geodesics for outgoing light rays and singular ones for incoming rays.
One interesting use of Eddington-Finkelstein coordinates is to investigate the radio pulses sent out by our astronaut colleague from Chapter 25 as she falls into the black hole. 13 13 ^(13){ }^{13}13

Example 26.7

Using outgoing Eddington-Finkelstein coordinates, the outgoing radial photons travel along lines of constant U U UUU, as shown in Fig. 26.7. The astronaut, whose proper time is τ τ tau\tauτ, undergoes a radial plunge with velocity u u u\boldsymbol{u}u with components u μ = ( u U , u r , 0 , 0 ) u μ = u U , u r , 0 , 0 u^(mu)=(u^(U),u^(r),0,0)u^{\mu}=\left(u^{U}, u^{r}, 0,0\right)uμ=(uU,ur,0,0). The astronaut emits regular light pulses which are received by an observer at infinity, whose proper time is t t ttt.
Let's start by finding the velocity components d U d τ = u U d U d τ = u U (dU)/((d)tau)=u^(U)\frac{\mathrm{d} U}{\mathrm{~d} \tau}=u^{U}dU dτ=uU and d r d τ = u r d r d τ = u r (dr)/((d)tau)=u^(r)\frac{\mathrm{d} r}{\mathrm{~d} \tau}=u^{r}dr dτ=ur. Notice how the metric doesn't depend on U U UUU, in exactly the manner in which it doesn't depend term in the Eddington-Finkelstein coordinates means we need to be slightly careful in dealing with up and down components. We have, expressing the ( U , r ) ( U , r ) (U,r)(U, r)(U,r) part of the metric in matrix form,
(26.28) g μ ν = ( ( 1 2 M r ) 1 1 0 ) , g μ ν = ( 0 1 1 ( 1 2 M r ) ) (26.28) g μ ν = 1 2 M r 1 1 0 , g μ ν = 0 1 1 1 2 M r {:[(26.28)g_(mu nu)=([-(1-(2M)/(r)),-1],[-1,0])","],[g^(mu nu)=([0,-1],[-1,(1-(2M)/(r))])]:}\begin{gather*} g_{\mu \nu}=\left(\begin{array}{cc} -\left(1-\frac{2 M}{r}\right) & -1 \\ -1 & 0 \end{array}\right), \tag{26.28}\\ g^{\mu \nu}=\left(\begin{array}{cc} 0 & -1 \\ -1 & \left(1-\frac{2 M}{r}\right) \end{array}\right) \end{gather*}(26.28)gμν=((12Mr)110),gμν=(011(12Mr))
whose inverse is
Fig. 26.6 Light pulses, shown in incoming Eddington-Finkelstein coordinates, sent outwards by an astronaut stood on the surface of a collapsing star.
13 13 ^(13){ }^{13}13 In this example, we follow the approach used in the book of problems by Lightman et al.
Fig. 26.7 An astronaut falling in the Schwarzschild geometry, emitting light pulses. These are shown in outgoing Eddington-Finkelstein coordinates where we set t = U + r t = U + r t=U+rt=U+rt=U+r and view the t r t r t-rt-rtr plane.
14 14 ^(14){ }^{14}14 Since t = U + r t = U + r t=U+r^(**)t=U+r^{*}t=U+r and the observations are made at the same position, we have Δ t = Δ U Δ t = Δ U Deltat_(oo)=DeltaU_(oo)\Delta t_{\infty}=\Delta U_{\infty}Δt=ΔU.
Normalization tells is that u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1, so we have
(26.30) g r r ( u r ) 2 + 2 g r U u U u r + g U U u U u U = 1 (26.30) g r r u r 2 + 2 g r U u U u r + g U U u U u U = 1 {:(26.30)g^(rr)(u_(r))^(2)+2g^(rU)u_(U)u_(r)+g^(UU)u_(U)u_(U)=-1:}\begin{equation*} g^{r r}\left(u_{r}\right)^{2}+2 g^{r U} u_{U} u_{r}+g^{U U} u_{U} u_{U}=-1 \tag{26.30} \end{equation*}(26.30)grr(ur)2+2grUuUur+gUUuUuU=1
from which, on substituting E ~ = u U E ~ = u U tilde(E)=-u_(U)\tilde{E}=-u_{U}E~=uU, we obtain
(26.31) ( 1 2 M r ) ( u r ) 2 + 2 E ~ u r + 1 = 0 (26.31) 1 2 M r u r 2 + 2 E ~ u r + 1 = 0 {:(26.31)(1-(2M)/(r))(u_(r))^(2)+2 tilde(E)u_(r)+1=0:}\begin{equation*} \left(1-\frac{2 M}{r}\right)\left(u_{r}\right)^{2}+2 \tilde{E} u_{r}+1=0 \tag{26.31} \end{equation*}(26.31)(12Mr)(ur)2+2E~ur+1=0
and so
(26.32) u r = E ~ [ E ~ 2 ( 1 2 M r ) ] 1 2 1 2 M r (26.32) u r = E ~ E ~ 2 1 2 M r 1 2 1 2 M r {:(26.32)u_(r)=(-( tilde(E))-[ tilde(E)^(2)-(1-(2M)/(r))]^((1)/(2)))/(1-(2M)/(r)):}\begin{equation*} u_{r}=\frac{-\tilde{E}-\left[\tilde{E}^{2}-\left(1-\frac{2 M}{r}\right)\right]^{\frac{1}{2}}}{1-\frac{2 M}{r}} \tag{26.32} \end{equation*}(26.32)ur=E~[E~2(12Mr)]1212Mr
We then have
(26.33) u U = g U U u U + g U r u r = 0 + E ~ + ( E ~ 2 1 + 2 M r ) 1 2 1 2 M r (26.33) u U = g U U u U + g U r u r = 0 + E ~ + E ~ 2 1 + 2 M r 1 2 1 2 M r {:(26.33)u^(U)=g^(UU)u_(U)+g^(Ur)u_(r)=0+(( tilde(E))+( tilde(E)^(2)-1+(2M)/(r))^((1)/(2)))/(1-(2M)/(r)):}\begin{equation*} u^{U}=g^{U U} u_{U}+g^{U r} u_{r}=0+\frac{\tilde{E}+\left(\tilde{E}^{2}-1+\frac{2 M}{r}\right)^{\frac{1}{2}}}{1-\frac{2 M}{r}} \tag{26.33} \end{equation*}(26.33)uU=gUUuU+gUrur=0+E~+(E~21+2Mr)1212Mr
and also
u r = g U r u U + g r r U r (26.34) = E ~ + [ E ~ ( E ~ 2 1 + 2 M r ) 1 2 ] = ( E ~ 2 1 + 2 M r ) 1 2 . u r = g U r u U + g r r U r (26.34) = E ~ + E ~ E ~ 2 1 + 2 M r 1 2 = E ~ 2 1 + 2 M r 1 2 . {:[u^(r)=g^(Ur)u_(U)+g^(rr)U_(r)],[(26.34)= tilde(E)+[-( tilde(E))-( tilde(E)^(2)-1+(2M)/(r))^((1)/(2))]=-( tilde(E)^(2)-1+(2M)/(r))^((1)/(2)).]:}\begin{align*} u^{r} & =g^{U r} u_{U}+g^{r r} U_{r} \\ & =\tilde{E}+\left[-\tilde{E}-\left(\tilde{E}^{2}-1+\frac{2 M}{r}\right)^{\frac{1}{2}}\right]=-\left(\tilde{E}^{2}-1+\frac{2 M}{r}\right)^{\frac{1}{2}} . \tag{26.34} \end{align*}ur=gUruU+grrUr(26.34)=E~+[E~(E~21+2Mr)12]=(E~21+2Mr)12.
This completes our analysis of the velocity in these coordinates.
Now consider the light pulses in Fig. 26.7. The astronaut sends (by her reckoning) regular pulses of light at intervals ( Δ τ ) em ( Δ τ ) em (Delta tau)_(em)(\Delta \tau)_{\mathrm{em}}(Δτ)em to the observer at infinity who detects them separated by intervals ( Δ t ) ( Δ t ) (Delta t)_(oo)(\Delta t)_{\infty}(Δt). We are interested in the ratio of wavelengths 14 14 ^(14){ }^{14}14
(26.35) λ λ em = ( Δ t ) ( Δ τ ) em = ( Δ U ) ( Δ τ ) em (26.35) λ λ em = ( Δ t ) ( Δ τ ) em = ( Δ U ) ( Δ τ ) em {:(26.35)(lambda_(oo))/(lambda_(em))=((Delta t)_(oo))/((Delta tau)_(em))=((Delta U)_(oo))/((Delta tau)_(em)):}\begin{equation*} \frac{\lambda_{\infty}}{\lambda_{\mathrm{em}}}=\frac{(\Delta t)_{\infty}}{(\Delta \tau)_{\mathrm{em}}}=\frac{(\Delta U)_{\infty}}{(\Delta \tau)_{\mathrm{em}}} \tag{26.35} \end{equation*}(26.35)λλem=(Δt)(Δτ)em=(ΔU)(Δτ)em
Now we receive the payoff for using these coordinates: since U U UUU is constant along the null geodesics then the change in U U UUU between pulses must be the same at the astronaut's position and at infinity ( or Δ U em = Δ U Δ U em = Δ U DeltaU_(em)=DeltaU_(oo)\Delta U_{\mathrm{em}}=\Delta U_{\infty}ΔUem=ΔU, see Fig. 26.7). We therefore have
(26.36) λ λ em = ( Δ U ) em ( Δ τ ) em = d U d τ | em (26.36) λ λ em = ( Δ U ) em ( Δ τ ) em = d U d τ em {:(26.36)(lambda_(oo))/(lambda_(em))=((Delta U)_(em))/((Delta tau)_(em))=(dU)/((d)tau)|_(em):}\begin{equation*} \frac{\lambda_{\infty}}{\lambda_{\mathrm{em}}}=\frac{(\Delta U)_{\mathrm{em}}}{(\Delta \tau)_{\mathrm{em}}}=\left.\frac{\mathrm{d} U}{\mathrm{~d} \tau}\right|_{\mathrm{em}} \tag{26.36} \end{equation*}(26.36)λλem=(ΔU)em(Δτ)em=dU dτ|em
The ratio is equal to the velocity component u U u U u^(U)u^{U}uU of the emitter. This is good progress, but is rather difficult to interpret. However, we can massage this into something more enlightening by using
(26.37) u r u U = d r d U = ( E ~ 2 1 + 2 M r ) 1 2 ( 1 2 M r ) E ~ + ( E ~ 2 1 + 2 M r ) 1 2 (26.37) u r u U = d r d U = E ~ 2 1 + 2 M r 1 2 1 2 M r E ~ + E ~ 2 1 + 2 M r 1 2 {:(26.37)(u^(r))/(u^(U))=(dr)/((d)U)=(-( tilde(E)^(2)-1+(2M)/(r))^((1)/(2))(1-(2M)/(r)))/(( tilde(E))+( tilde(E)^(2)-1+(2M)/(r))^((1)/(2))):}\begin{equation*} \frac{u^{r}}{u^{U}}=\frac{\mathrm{d} r}{\mathrm{~d} U}=\frac{-\left(\tilde{E}^{2}-1+\frac{2 M}{r}\right)^{\frac{1}{2}}\left(1-\frac{2 M}{r}\right)}{\tilde{E}+\left(\tilde{E}^{2}-1+\frac{2 M}{r}\right)^{\frac{1}{2}}} \tag{26.37} \end{equation*}(26.37)uruU=dr dU=(E~21+2Mr)12(12Mr)E~+(E~21+2Mr)12
This latter expression can be expanded near r = 2 M r = 2 M r=2Mr=2 Mr=2M to yield the shift in the pulses emitted close to the horizon. We obtain
so that
(26.38) d U 2 ( 1 2 M r ) 1 d r 2 ( r 2 M 1 ) 1 d r (26.38) d U 2 1 2 M r 1 d r 2 r 2 M 1 1 d r {:(26.38)dU~~-2(1-(2M)/(r))^(-1)dr~~-2((r)/(2M)-1)^(-1)dr:}\begin{equation*} \mathrm{d} U \approx-2\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r \approx-2\left(\frac{r}{2 M}-1\right)^{-1} \mathrm{~d} r \tag{26.38} \end{equation*}(26.38)dU2(12Mr)1 dr2(r2M1)1 dr
or
(26.39) U 4 M ln ( r 2 M 1 ) + const (26.39) U 4 M ln r 2 M 1 +  const  {:(26.39)U~~-4M ln((r)/(2M)-1)+" const ":}\begin{equation*} U \approx-4 M \ln \left(\frac{r}{2 M}-1\right)+\text { const } \tag{26.39} \end{equation*}(26.39)U4Mln(r2M1)+ const 
1 2 M r e U 4 M . 1 2 M r e U 4 M . 1-(2M)/(r)~~e^(-(U)/(4M)).1-\frac{2 M}{r} \approx \mathrm{e}^{-\frac{U}{4 M}} .12MreU4M.
From which we predict from eqn 26.33, that in the limit r 2 M r 2 M r rarr2Mr \rightarrow 2 Mr2M we have
u U e U / 4 M . u U e U / 4 M . u^(U)prope^(U//4M).u^{U} \propto \mathrm{e}^{U / 4 M} .uUeU/4M.
ver at large r we have U t + const and so  ver at large  r  we have  U t +  const and so  " ver at large "r" we have "U~~t+" const and so "\text { ver at large } r \text { we have } U \approx t+\text { const and so } ver at large r we have Ut+ const and so 
(26.42) ω ω em = λ em λ e t d M (26.42) ω ω em = λ em λ e t d M {:(26.42)(omega_(oo))/(omega_(em))=(lambda_(em))/(lambda_(oo))~~e^(-(t)/(dM)):}\begin{equation*} \frac{\omega_{\infty}}{\omega_{\mathrm{em}}}=\frac{\lambda_{\mathrm{em}}}{\lambda_{\infty}} \approx \mathrm{e}^{-\frac{t}{d M}} \tag{26.42} \end{equation*}(26.42)ωωem=λemλetdM
We conclude that the light at infinity is redshifted: the detected frequency ω ω omega_(oo)\omega_{\infty}ω drops to zero exponentially with a time constant 4 M 4 M 4M4 M4M, that depends on the mass of the black hole.
So, from the point of view of the observer at infinity, not only does it take an infinite interval in coordinate time for the astronaut to fall through the event horizon, the light that comes from the observer is redshifted, eventually by an infinite amount. We, while sat safely at infinity, would see the astronaut simultaneously slowing and fading as she approaches the Schwarzschild radius.
Although useful, Eddington-Finkelstein coordinates do not provide a solution to the central problem of this chapter: finding a coordinate system to describe the whole of the Schwarzschild geometry. That solution is the subject of the next chapter.

Chapter summary

  • The Schwarzschild black hole has a physical singularity located at r = 0 r = 0 r=0r=0r=0, whose presence gives rise to closed trapped surfaces. It has a coordinate singularity at r S r S r_(S)r_{\mathrm{S}}rS that can be removed by a change in coordinates.
  • Eddington-Finkelstein coordinates remove the coordinate singularity for either incoming or outgoing rays, but not for both.

Exercises

26.1) Show that for the Schwarzschild metric we have
(26.43) R μ ν α β R μ ν α β = 12 M 2 r 6 (26.43) R μ ν α β R μ ν α β = 12 M 2 r 6 {:(26.43)R^(mu nu alpha beta)R_(mu nu alpha beta)=(12M^(2))/(r^(6)):}\begin{equation*} R^{\mu \nu \alpha \beta} R_{\mu \nu \alpha \beta}=\frac{12 M^{2}}{r^{6}} \tag{26.43} \end{equation*}(26.43)RμναβRμναβ=12M2r6
Use this to discuss the nature of the singularities at (i) r = 2 M r = 2 M r=2Mr=2 Mr=2M and (ii) r = 0 r = 0 r=0r=0r=0.
(26.2) Show that the Eddington-Finkelstein metric has determinant g = r 4 sin 2 θ g = r 4 sin 2 θ g=-r^(4)sin^(2)thetag=-r^{4} \sin ^{2} \thetag=r4sin2θ.
(26.3) Verify eqn 26.29 .
(26.4) In example 26.3, we guessed that U ( u ) = e u U ( u ) = e u U(u)=e^(-u)U(u)=\mathrm{e}^{-u}U(u)=eu and V ( v ) = e v V ( v ) = e v V(v)=e^(v)V(v)=\mathrm{e}^{v}V(v)=ev. The guesswork isn't necessary: there's a method we can use by considering how the affine parameter λ λ lambda\lambdaλ must vary along the null geodesics. (a) Show that we can write a constant of the mo-
tion
(26.44) E ~ = x 2 d t d λ (26.44) E ~ = x 2 d t d λ {:(26.44) tilde(E)=x^(2)((d)t)/((d)lambda):}\begin{equation*} \tilde{E}=x^{2} \frac{\mathrm{~d} t}{\mathrm{~d} \lambda} \tag{26.44} \end{equation*}(26.44)E~=x2 dt dλ
(b) Setting u u uuu constant, show that this leads to
(26.45) λ = C + ( e u 2 E ~ ) e v (26.45) λ = C + e u 2 E ~ e v {:(26.45)lambda=C+((e^(-u))/(2( tilde(E))))e^(v):}\begin{equation*} \lambda=C+\left(\frac{\mathrm{e}^{-u}}{2 \tilde{E}}\right) \mathrm{e}^{v} \tag{26.45} \end{equation*}(26.45)λ=C+(eu2E~)ev
with C C CCC a constant, showing that λ out = e v λ out  = e v lambda_("out ")=e^(v)\lambda_{\text {out }}=\mathrm{e}^{v}λout =ev is an affine parameter along the outgoing geodesics. (c) Show that a similar calculation yields λ in = λ in  = lambda_("in ")=\lambda_{\text {in }}=λin = e u e u -e^(-u)-\mathrm{e}^{-u}eu for the incoming geodesics.
We can then reparametrize the null coordinates u u uuu and v v vvv using the affine coordinates in the hope of simplifying them.

27

27.1 Enter the Kruskal metric 280 27.2 Wormholes 284 27.3 Another Penrose diagram Chapter summary 287 Exercises
Exercises 287
1 1 ^(1){ }^{1}1 Martin Kruskal (1925-2006) made several contributions across a wide range of subjects in mathematical physics and was especially noted for his work on solitons.
2 2 ^(2){ }^{2}2 In addition to George Szekeres' (19112005) contribution to relativity was his work with Esther Klein on an important problem in combinatorial geomtant problem in combinatorial geom-
etry. This was dubbed the 'HappyEnding problem' by Paul Erdős after Ending problem' by Paul Erdös after
the collaboration resulted in Szekeres the collaboration result
and Klein's marriage.
3 3 ^(3){ }^{3}3 The term r 2 d Ω 2 r 2 d Ω 2 r^(2)dOmega^(2)r^{2} \mathrm{~d} \Omega^{2}r2 dΩ2 is unchanged by our various manipulations and so we drop various manipulations and so we drop
it in the next example, before restoring it in the next exam
it in the solution.

Kruskal-Szekeres

coordinates
t in the solution.
And to add a new awareness to what the sky possesses in its size and formlessness, there is involved the quality of decay. For all the wonder of these everlasting stars, eternal spheres, and what not, they are not everlasting, they are not eternal; they burn out like candles.
Thomas Hardy (1840-1928) Two on a Tower
In this chapter, we introduce the solution to the main problem of the last chapter: defining a set of coordinates that are continuous over the event horizon and intuitively useful. These are the Kruskal 1 1 ^(1)-{ }^{1}-1 Szekeres 2 2 ^(2)^{2}2 coordinates.

27.1 Enter the Kruskal metric

The problem with Eddington-Finkelstein coordinates is that they represent two coordinate systems, neither satisfactory. For the outgoing version, incoming light rays are discontinuous at r S r S r_(S)r_{\mathrm{S}}rS (and vice-versa). Recall that, in the last chapter, we paused our programme of removing the singularity at r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M at the point of having massaged the Schwarzschild metric into the form
(27.1) d s 2 = ( 1 2 M r ) d U d V + r 2 d Ω 2 (27.1) d s 2 = 1 2 M r d U d V + r 2 d Ω 2 {:(27.1)ds^(2)=-(1-(2M)/(r))dUdV+r^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\left(1-\frac{2 M}{r}\right) \mathrm{d} U \mathrm{~d} V+r^{2} \mathrm{~d} \Omega^{2} \tag{27.1} \end{equation*}(27.1)ds2=(12Mr)dU dV+r2 dΩ2
where 3 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 3 d Ω 2 = d θ 2 + sin 2 θ d ϕ 2 ^(3)dOmega^(2)=dtheta^(2)+sin^(2)thetadphi^(2){ }^{3} \mathrm{~d} \Omega^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}3 dΩ2=dθ2+sin2θ dϕ2. We will now complete the process of removing this singularity.
Example 27.1
Use the fact that r = r + 2 M ln | r / 2 M 1 | = ( V U ) / 2 r = r + 2 M ln | r / 2 M 1 | = ( V U ) / 2 r^(**)=r+2M ln |r//2M-1|=(V-U)//2r^{*}=r+2 M \ln |r / 2 M-1|=(V-U) / 2r=r+2Mln|r/2M1|=(VU)/2 and rewrite the twodimensional version of the metric as
(27.2) d s 2 = ( 2 M r e r 2 M ) e ( V U ) 4 M d U d V (27.2) d s 2 = 2 M r e r 2 M e ( V U ) 4 M d U d V {:(27.2)ds^(2)=-((2M)/(r)e^(-(r)/(2M)))e^(((V-U))/(4M))dUdV:}\begin{equation*} \mathrm{d} s^{2}=-\left(\frac{2 M}{r} \mathrm{e}^{-\frac{r}{2 M}}\right) \mathrm{e}^{\frac{(V-U)}{4 M}} \mathrm{~d} U \mathrm{~d} V \tag{27.2} \end{equation*}(27.2)ds2=(2Mrer2M)e(VU)4M dU dV
The message here is that we have factorized the metric into a function of r r rrr that is non-singular as r 2 M r 2 M r rarr2Mr \rightarrow 2 Mr2M, by (i) an exponential function of U U UUU and V V VVV and (ii) the Minkowski light-cone metric d U d V d U d V -dUdV-\mathrm{d} U \mathrm{~d} VdU dV. Following exactly the same path as we did in the Rindler case in the last chapter, we transform away the exponential dependence on U U UUU and V V VVV using a change in coordinates
u = e U / 4 M u = e U / 4 M u=-e^(-U//4M)u=-\mathrm{e}^{-U / 4 M}u=eU/4M
(27.3) v = e V / 4 M (27.3) v = e V / 4 M {:(27.3)v=e^(V//4M):}\begin{equation*} v=\mathrm{e}^{V / 4 M} \tag{27.3} \end{equation*}(27.3)v=eV/4M
Fig. 27.1 Kruskal spacetime. The thick lines show the physical singularities at r = 0 r = 0 r=0r=0r=0. Regions I and II are the parts through which a collapsing star moves. Regions I' and II' are described in Section 27.2.
This gives a metric
(27.4) d s 2 = 32 M 3 e r / 2 M r d u d u (27.4) d s 2 = 32 M 3 e r / 2 M r d u d u {:(27.4)ds^(2)=-(32M^(3)e^(-r//2M))/(r)dudu:}\begin{equation*} \mathrm{d} s^{2}=-\frac{32 M^{3} \mathrm{e}^{-r / 2 M}}{r} \mathrm{~d} u \mathrm{~d} u \tag{27.4} \end{equation*}(27.4)ds2=32M3er/2Mr du du
Once again, we can make the (now familiar) final transformation
T = ( u + v ) / 2 , (27.5) X = ( v u ) / 2 . T = ( u + v ) / 2 , (27.5) X = ( v u ) / 2 . {:[T=(u+v)//2","],[(27.5)X=(v-u)//2.]:}\begin{align*} & T=(u+v) / 2, \\ & X=(v-u) / 2 . \tag{27.5} \end{align*}T=(u+v)/2,(27.5)X=(vu)/2.
From the argument in the last example we obtain 4 4 ^(4){ }^{4}4 the Kruskal metric
(27.6) d s 2 = 32 M 3 e r / 2 M r ( d T 2 + d X 2 ) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (27.6) d s 2 = 32 M 3 e r / 2 M r d T 2 + d X 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(27.6)ds^(2)=(32M^(3)e^(-r//2M))/(r)(-dT^(2)+dX^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=\frac{32 M^{3} \mathrm{e}^{-r / 2 M}}{r}\left(-\mathrm{d} T^{2}+\mathrm{d} X^{2}\right)+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{27.6} \end{equation*}(27.6)ds2=32M3er/2Mr(dT2+dX2)+r2( dθ2+sin2θ dϕ2)
The Kruskal metric is not singular at r = 2 M r = 2 M r=2Mr=2 Mr=2M, demonstrating that this point is indeed a coordinate singularity, as we had previously claimed. The metric's structure is represented in Fig. 27.1 in the X T X T X-TX-TXT plane with two coordinates suppressed, in a so-called Kruskal diagram. (Suppressing two coordinates means that each point in the Kruskal diagram represents a two-dimensional sphere of radius r r rrr.) By construction, all null geodesics are 45 45 45^(@)45^{\circ}45 lines, not just half of them as we had in EddingtonFinkelstein coordinates. Therefore, the light-cone structure is that of Minkowski space with future light cones heading towards positive T T TTT, and past light cones heading towards negative T T TTT. Inward-directed rays head towards small X X XXX, outward-directed rays head towards large X X XXX. Note that the metric becomes flat, asymptotically, in the limit X X X rarr ooX \rightarrow \inftyX. Let's look at some of the features of the Kruskal diagram.
4 4 ^(4){ }^{4}4 The relation between old and new coordinates is
( 1 r 2 M ) e r / 2 M = T 2 X 2 1 r 2 M e r / 2 M = T 2 X 2 (1-(r)/(2M))e^(r//2M)=T^(2)-X^(2)\left(1-\frac{r}{2 M}\right) \mathrm{e}^{r / 2 M}=T^{2}-X^{2}(1r2M)er/2M=T2X2, t 4 M = tanh 1 ( T X ) t 4 M = tanh 1 T X (t)/(4M)=tanh^(-1)((T)/(X))\frac{t}{4 M}=\tanh ^{-1}\left(\frac{T}{X}\right)t4M=tanh1(TX).
Fig. 27.2 Kruskal spacetime with lines of constant r r rrr plotted. They obey T 2 T 2 T^(2)-T^{2}-T2 X 2 = ( 1 r / 2 M ) e r / 2 M X 2 = ( 1 r / 2 M ) e r / 2 M X^(2)=(1-r//2M)e^(r//2M)X^{2}=(1-r / 2 M) \mathrm{e}^{r / 2 M}X2=(1r/2M)er/2M.
Fig. 27.3 Kruskal spacetime with (straight) lines of constant t t ttt plotted. They obey T / X = tanh ( t / 4 M ) T / X = tanh ( t / 4 M ) T//X=tanh(t//4M)T / X=\tanh (t / 4 M)T/X=tanh(t/4M) in region I and X / T = tanh ( t / 4 M ) X / T = tanh ( t / 4 M ) X//T=tanh(t//4M)X / T=\tanh (t / 4 M)X/T=tanh(t/4M) in region II.
5 5 ^(5){ }^{5}5 If you're tempted to give up at this point, remember the motivation: these point, remember the motivation: these
coordinates ultimately make things as coordinates ultimately make things as
simple as possible, not least as they remove the coordinate singularity evident in the Schwarzschild coordinates at r S r S r_(S)r_{\mathrm{S}}rS. In the words of J. L. Synge, "If the problem of spherical symmetry had been attacked originally in this way, it would never have occurred to anyone to think of a singularity here."
Fig. 27.4 Kruskal diagram showing an in-falling timelike world line.
Fig. 27.5 Kruskal diagram showing light pulses emitted by an astronaut falling rapidly into a black hole, received by an observer at a fixed radius.
  • In the Kruskal diagram, lines of constant r r rrr are curves of constant X 2 T 2 [ = ( 1 r / 2 M ) e r / 2 M ] X 2 T 2 = ( 1 r / 2 M ) e r / 2 M X^(2)-T^(2)[=-(1-r//2M)e^(r//2M)]X^{2}-T^{2}\left[=-(1-r / 2 M) \mathrm{e}^{r / 2 M}\right]X2T2[=(1r/2M)er/2M] and are therefore hyperbolae, as shown in Fig. 27.2. One notable hyperbola is found for r = 2 M r = 2 M r=2Mr=2 Mr=2M, for which X 2 T 2 = 0 X 2 T 2 = 0 X^(2)-T^(2)=0X^{2}-T^{2}=0X2T2=0, resulting in the lines X = ± T X = ± T X=+-TX= \pm TX=±T, which cut the plane into four regions, labelled I, II, I' and II' on the diagram in Fig. 27.1. Another notable hyperbola is that for r = 0 r = 0 r=0r=0r=0, which is a physical singularity shown conventionally by thick (or sometimes jagged) lines which represent X 2 T 2 = 1 X 2 T 2 = 1 X^(2)-T^(2)=-1X^{2}-T^{2}=-1X2T2=1. There are two r = 0 r = 0 r=0r=0r=0 singularities: one in the past (at the bottom of the diagram) and one in the future (at the top).
  • The spacetime covered by Schwarzschild coordinates r > 0 r > 0 r > 0r>0r>0 and < t < < t < -oo < t < oo-\infty<t<\infty<t< is represented by that half of the Kruskal diagram with T > X T > X T > -XT>-XT>X, comprising regions I and II. Schwarzschild coordinates outside of the event horizon ( 2 M < r < ( 2 M < r < (2M < r < oo(2 M<r<\infty(2M<r< and < t < ) < t < ) -oo < t < oo)-\infty<t<\infty)<t<) cover region I with X > 0 X > 0 X > 0X>0X>0 and X < T < X X < T < X -X < T < X-X<T<XX<T<X. In contrast, region II covers that part within the event horizon ( 0 < r < 2 M ( 0 < r < 2 M (0 < r < 2M(0<r<2 M(0<r<2M and < t < ) < t < ) -oo < t < oo)-\infty<t<\infty)<t<) where T > 0 T > 0 T > 0T>0T>0 and T < X < T T < X < T -T < X < T-T<X<TT<X<T.
  • Spacelike hypersurfaces of constant t t ttt are represented by lines of constant T / X T / X T//XT / XT/X and so are straight lines, as shown in Fig. 27.3. The line corresponding to the surface t = 0 t = 0 t=0t=0t=0 has T = 0 T = 0 T=0T=0T=0 for r > 2 M r > 2 M r > 2Mr>2 Mr>2M, but X = 0 X = 0 X=0X=0X=0 for r < 2 M r < 2 M r < 2Mr<2 Mr<2M. The coordinate t = t = t=oot=\inftyt= is represented by X = T X = T X=TX=TX=T and the coordinate t = t = t=-oot=-\inftyt= is represented by X = T X = T X=-TX=-TX=T. In each quadrant, the t t ttt coordinate advances between t = t = t=-oot=-\inftyt= and oo\infty.
Clearly, these coordinates take a little getting used to. 5 5 ^(5){ }^{5}5 However, the fact that the light rays all travel at 45 45 45^(@)45^{\circ}45 to the vertical simplifies things nicely. Let's look at an example of using the coordinates.

Example 27.2

The path of the in-falling astronaut is shown in a Kruskal diagram in Fig. 27.4. The world line shows the astronaut passing from region I, through the horizon into region II, and then falling towards the singularity at r = 0 r = 0 r=0r=0r=0. Remember that light rays travel at 45 45 45^(@)45^{\circ}45, so on passing the r = 2 M r = 2 M r=2Mr=2 Mr=2M horizon, all light rays emitted along the astronaut's path will end up hitting the r = 0 r = 0 r=0r=0r=0 singularity or, if you like, they fall into the hole. An illustration of the light signals emitted by the in-falling victim is shown in Fig. 27.5. The observer in region I remains at a fixed value of r r rrr (presumably maintained by firing a rocket, or some such) and their world line is therefore a hyperbola. The astronaut falls into the hole (rather fast, as shown, although the angle is less The astronaut falls into the hole (rather fast, as shown, although the angle is less
than 45 45 45^(@)45^{\circ}45 to maintain a subluminal speed). As the astronaut falls they regularly emit light pulses. Since the observer follows a hyperbolic path on the Kruskal diagram, we see that the observer receives the signals less and less frequently. The last ones are received after long intervals of coordinate time, with the ones emitted just above the horizon not reaching the observer until t t t rarr oot \rightarrow \inftyt. Pulses emitted after the astronaut has fallen into the hole are seen to hit the singularity at r = 0 r = 0 r=0r=0r=0, as indeed the astronaut does, where they are destroyed.
It is immediately apparent that Schwarzschild coordinates only cover regions I and II of the Kruskal diagram, and so the other half-plane
(with T < X T < X T < -XT<-XT<X ) doesn't yet seem of much physical significance. However, it is possible to interpret the whole of the X T X T X-TX-TXT plane as a meaningful coordinate patch. Taking all of the Kruskal coordinates to describe spacetime involves the extension of Schwarzschild space into the T < X T < X T < -XT<-XT<X half plane (the previously mysterious regions I' and II' in Fig. 27.1), and is therefore known as the Kruskal extension. Before exploring this feature, let's try to make ourselves a little more comfortable with the Kruskal coordinates in the next example.
Example 27.3
We can understand the physical basis of the (still admittedly rather mysterious) Kruskal coordinates with an analogy made by Wolfgang Rindler. We consider the acceleration of a particle in Minkowski space using coordinates ( T , X , Y , Z T , X , Y , Z T,X,Y,ZT, X, Y, ZT,X,Y,Z ). The particle's world line is X 2 T 2 = x 2 X 2 T 2 = x 2 X^(2)-T^(2)=x^(2)X^{2}-T^{2}=x^{2}X2T2=x2, which corresponds to a proper acceleration in the X X XXX direction of 1 / x 1 / x 1//x1 / x1/x. [This problem is discussed in Chapter 2. Some example hyperbolic world lines are shown in Fig. 27.6(a).] Now consider a change of variables to a set of coordinates ( t , x , y , z ) ( t , x , y , z ) (t,x,y,z)(t, x, y, z)(t,x,y,z) where
T = x sinh t , X = x cosh t , Y = y , Z = z T = x sinh t , X = x cosh t , Y = y , Z = z T=x sinh t,quad X=x cosh t,quad Y=y,quad Z=zT=x \sinh t, \quad X=x \cosh t, \quad Y=y, \quad Z=zT=xsinht,X=xcosht,Y=y,Z=z
(27.7)
This transformation itself obeys the world-line equation for an accelerating particle and so ( t , x , y , z t , x , y , z t,x,y,zt, x, y, zt,x,y,z ) represent an accelerating set of coordinates. 6 6 ^(6){ }^{6}6 Lines of constant x x xxx, when graphed in region I of the ( T , X , Y , Z ) ( T , X , Y , Z ) (T,X,Y,Z)(T, X, Y, Z)(T,X,Y,Z) coordinate system on a Minkowski diagram in the T T TTT - X X XXX plane, give hyperbolic curves [Fig. 27.6(b)]. 7 7 ^(7){ }^{7}7 So, it is as if the coordinates ( t , x , y , z ) ( t , x , y , z ) (t,x,y,z)(t, x, y, z)(t,x,y,z) are the coordinates used inside a rocket that accelerates in the X X XXX-direction. (By the same token, region I' corresponds to a rocket accelerating in the X X -X-XX direction.)
In order to describe regions II and II' in the same way, we make the transformation
(27.8) T = x cosh t , X = x sinh t , Y = y , Z = z , (27.8) T = x cosh t , X = x sinh t , Y = y , Z = z , {:(27.8)T=x cosh t","quad X=x sinh t","quad Y=y","quad Z=z",":}\begin{equation*} T=x \cosh t, \quad X=x \sinh t, \quad Y=y, \quad Z=z, \tag{27.8} \end{equation*}(27.8)T=xcosht,X=xsinht,Y=y,Z=z,
which obeys X 2 T 2 = x 2 X 2 T 2 = x 2 X^(2)-T^(2)=-x^(2)X^{2}-T^{2}=-x^{2}X2T2=x2, showing an exchange in roles of the T T TTT and X X XXX coordinates in these regions. The resulting accelerating world lines in regions I and II are shown in Fig. 27.6(b).
Using the ( t , x , y , z t , x , y , z t,x,y,zt, x, y, zt,x,y,z ) coordinate system in regions I and I' causes the Minkowski metric to change to the Rindler metric
(27.9) d s 2 = x 2 d t 2 + d x 2 + d y 2 + d z 2 (27.9) d s 2 = x 2 d t 2 + d x 2 + d y 2 + d z 2 {:(27.9)ds^(2)=-x^(2)dt^(2)+dx^(2)+dy^(2)+dz^(2):}\begin{equation*} \mathrm{d} s^{2}=-x^{2} \mathrm{~d} t^{2}+\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2} \tag{27.9} \end{equation*}(27.9)ds2=x2 dt2+dx2+dy2+dz2
while in regions II and II' we have
(27.10) d s 2 = x 2 d t 2 d x 2 + d y 2 + d z 2 (27.10) d s 2 = x 2 d t 2 d x 2 + d y 2 + d z 2 {:(27.10)ds^(2)=x^(2)dt^(2)-dx^(2)+dy^(2)+dz^(2):}\begin{equation*} \mathrm{d} s^{2}=x^{2} \mathrm{~d} t^{2}-\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2} \tag{27.10} \end{equation*}(27.10)ds2=x2 dt2dx2+dy2+dz2
If we then make the change of variables x 2 = ( 2 r 1 ) x 2 = ( 2 r 1 ) x^(2)=(2r-1)x^{2}=(2 r-1)x2=(2r1) in regions I and I' and x 2 = ( 2 r 1 ) x 2 = ( 2 r 1 ) -x^(2)=(2r-1)-x^{2}=(2 r-1)x2=(2r1) in regions II and II', then we can describe the whole of Minkowski space with the accelerating coordinates and a metric
(27.11) d s 2 = ( 2 r 1 ) d t 2 + ( 2 r 1 ) 1 d x 2 + d y 2 + d z 2 (27.11) d s 2 = ( 2 r 1 ) d t 2 + ( 2 r 1 ) 1 d x 2 + d y 2 + d z 2 {:(27.11)ds^(2)=-(2r-1)dt^(2)+(2r-1)^(-1)dx^(2)+dy^(2)+dz^(2):}\begin{equation*} \mathrm{d} s^{2}=-(2 r-1) \mathrm{d} t^{2}+(2 r-1)^{-1} \mathrm{~d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2} \tag{27.11} \end{equation*}(27.11)ds2=(2r1)dt2+(2r1)1 dx2+dy2+dz2
This previous equation rather closely resembles the Schwarzschild metric line element! So the accelerating coordinate system (used as a natural choice inside a rocket) can be likened to the Schwarzschild system, while Minkowski space is analogous to the Kruskal system. We therefore identify the accelerating coordinate system ( t , x , y , z ) ( t , x , y , z ) (t,x,y,z)(t, x, y, z)(t,x,y,z) with Schwarzschild coordinates and the Minkowski space [with coordinates ( T , X , Y , Z ) ] T , X , Y , Z ) ] T,X,Y,Z)]T, X, Y, Z)]T,X,Y,Z)] with Kruskal space and we have the analogy:
(27.12) ( Minkowski coordinates Kruskal coordinates ) ( Accelerating Minkowski coordinates ( Schwarzschild coordinates ) ) (27.12)  Minkowski   coordinates   Kruskal   coordinates   Accelerating Minkowski   coordinates  (  Schwarzschild   coordinates  ) {:(27.12)([" Minkowski "],[" coordinates "],[" Kruskal "],[" coordinates "])longleftrightarrow([" Accelerating Minkowski "],[" coordinates "],[((" Schwarzschild ")/(" coordinates "))]):}\left(\begin{array}{c} \text { Minkowski } \tag{27.12}\\ \text { coordinates } \\ \text { Kruskal } \\ \text { coordinates } \end{array}\right) \longleftrightarrow\left(\begin{array}{c} \text { Accelerating Minkowski } \\ \text { coordinates } \\ \binom{\text { Schwarzschild }}{\text { coordinates }} \end{array}\right)(27.12)( Minkowski  coordinates  Kruskal  coordinates )( Accelerating Minkowski  coordinates ( Schwarzschild  coordinates ))
We can use the analogy to say that, in region I, analogous to the accelerating rocket system in a Minkowski coordinate system, is a Schwarzschild system held outside r = r S r = r S r=r_(S)r=r_{\mathrm{S}}r=rS represented in the Kruskal system.
6 6 ^(6){ }^{6}6 Specifically, each choice of x x xxx represents the path of an accelerating particle in the ( T , X , Y , Z T , X , Y , Z T,X,Y,ZT, X, Y, ZT,X,Y,Z ) system, with acceleration 1 / x 1 / x 1//x1 / x1/x. Region I of the T X T X T-XT-XTX plane can also be used to represent the plane can also be used to represent the
acceleration of a rigid body. Measured in the ( T , X , Y , Z ) ( T , X , Y , Z ) (T,X,Y,Z)(T, X, Y, Z)(T,X,Y,Z) system, we see from the horizontal bars in Fig. 27.6(a) that if the front of a rocket has some acceleration a = 1 / x a = 1 / x a=1//xa=1 / xa=1/x, then the back must have greater acceleration leading to a increasing length contraction.
7 7 ^(7){ }^{7}7 Since we also have X / T = coth t X / T = coth t X//T=coth tX / T=\operatorname{coth} tX/T=cotht, lines of constant t t ttt are straight in the T X T X T-XT-XTX plane.
Fig. 27.6 (a) Hyperbolae in region I showing accelerating particles in Minkowski space. Horizontal lines linking two coordinates values of x x xxx at fixed T T TTT show how the back of a rigid object appears to accelerate faster than the front. (b) The accelerating coordinate system in regions I (constant x x xxx ) and II (constant t t ttt ). The resulting picture resembles the Kruskal diagram.
Fig. 27.7 A timelike slice of Kruskal spacetime taken at T = 0 T = 0 T=0T=0T=0. This shows the spacetime throat between the universe I (top) and universe I' (bottom).
The last example shows that rather than regarding the Kruskal coordinates as a complicated and contrived system, we should perhaps think of them as being as natural a system as the Minkowski coordinates are in flat space. The Schwarzschild coordinates are then like the 'difficult' coordinates of an accelerating frame.

27.2 Wormholes

Let's now return to the idea of taking the Kruskal system to cover all of spacetime, previously named the Kruskal extension. Intriguingly, the Kruskal extension describes the spacetime of a wormhole. Large positive values of X X XXX in region I corresponds to flat spacetime and so, in the same way, large, negative X X XXX in region I' also corresponds to a flat spacetime. Region I' therefore looks very much like the spacetime for a spherical mass, outside the Schwarzschild radius. It is therefore interpreted as another universe outside of a black hole. Since time increases up the page, region II' is that part of spacetime that some past-directed world lines will fall into from region I' (just as future directed light rays can fall from region I into the black hole in region II). This implies that the primed universe (regions I' and I") is time-reversed compared to the universe comprising regions I and II, which is to say that the coordinate t t ttt advances in the opposite sense. Region II' has its own r = 0 r = 0 r=0r=0r=0 singularity and so any particle in region II' must have been 'created' at this past singularity, just as any particle in region II must be destroyed at the future singularity. From this point of view, the past singularity in region II' might be characterized as a white hole from which all particles and light in region II' must originate.
The two singular r = 0 r = 0 r=0r=0r=0 lines in regions II and II' form a throat in spacetime separating the two asymptotically flat universes in regions I and I'. A spacelike hypersurface at T = 0 T = 0 T=0T=0T=0 is shown in Fig. 27.7 where the throat, of radius 2 M 2 M 2M2 M2M at this point is shown.
Is it possible to travel between the two universes? That is, can a traveller traverse the throat in Fig. 27.7? The short answer is no: there are no null or timelike geodesics linking regions I and I', and so all photons and massive particles crossing the line r = 2 M r = 2 M r=2Mr=2 Mr=2M are doomed to fall towards a singularity. This can be seen in Fig. 27.1, where we can see that no non-spacelike (i.e. 45 45 <= 45^(@)\leq 45^{\circ}45 to the vertical) curve can get from region I to region I'. (Notice how the Kruskal coordinates, now proving their worth, allow this to be seen relatively simply.) Despite the impossibility of a trip between universes, there are some intriguing features connected with the wormhole that we examine in the next example.
Fig. 27.8 Evolution of timelike slices of Kruskal spacetime with T T TTT.
Example 27.4
We usually regard the Schwarzschild geometry as static, with increments δ t δ t delta t\delta tδt in time t t ttt causing no change in the geometry. This enables us to identify spacelike hypersurfaces, such as that shown in Fig. 27.7 for T = 0 T = 0 T=0T=0T=0, showing a throat linking the two asymptotically flat universes. However, the idea of the static metric only makes sense asymptotically flat universes. However, the idea of the static metric only makes sense
in regions I and I'. In regions II and II', increments in the coordinate t t ttt are actually in regions I and I'. In regions II and II', increments in the coordinate t t ttt are actually
increments in space and so spacelike hypersurfaces are not really static. Instead, they increments in space and so spacelike hypersurfaces are not really static. Instead, they
can be thought of as evolving in the increasing T T TTT direction (i.e. the future, according can be thought of as evolving in the increasing T T TTT direction (i.e. the future, according
to light cones). From this point of view we can trace the evolution of an arbitrary spacelike slice 8 8 ^(8){ }^{8}8 as a function of T T TTT.
Consider, for example, the spacelike slice A A A A A^(')-AA^{\prime}-AAA show in Fig. 27.8. We see (right) that the two universes are unlinked. Each slice curves towards the singularity at r = 0 r = 0 r=0r=0r=0 represented by a cusp, which occurs where the slice hits the singularity. At a larger value of T T TTT, when the slice just touches the singularity, the universes must meet at a pinch point and then, as T T TTT increases, the universes are linked by a throat that avoids r = 0 r = 0 r=0r=0r=0. The throat takes its maximum size of r = 2 M r = 2 M r=2Mr=2 Mr=2M at T = 0 ( B B ) T = 0 B B T=0(B^(')-B)T=0\left(B^{\prime}-B\right)T=0(BB) before contracting. The universes then reach a pinch point again before separating ( C C C C C^(')-CC^{\prime}-CCC ). contracting. The universes then reach a pinch point again before separating ( C = C C = C C=-CC=-CC=C ). Although it might look temptingly like a particle or signal could be sent from region
I, through the throat of the wormhole, to region I', it cannot. From the point of I, through the throat of the wormhole, to region I', it cannot. From the point of
view of the spacelike surfaces, the particle cannot move fast enough to traverse the view of the spacelike surfaces, the particle cannot move fast enough to traverse the
throat. The spacelike slices of the spacetime evolve so fast that the universes pinch off before the signal (or traveller) has reached the middle of the throat. The signal is doomed to hit the r = 0 r = 0 r=0r=0r=0 singularity where it will be destroyed.

27.3 Another Penrose diagram

The Penrose diagram for the Kruskal extension is shown in Fig. 27.9(a). It shows regions I, II, I' and II'. There are two spacelike singularities, one
8 8 ^(8){ }^{8}8 There is no meaningful, special spacelike slice we can take. We simply select an arbitrary one to show the evolution.
Fig. 27.9 (a) Penrose diagram of the Kruskal extension. (b) The same, but showing lines of constant r r rrr and t t ttt.
in the past and one in the future. Notice how these exist at the edges of the diagram or, if you prefer, the edges of spacetime. Unlike the case of the Robertson-Walker spacetimes in Chapter 19, not all world lines meet the singularities. In fact there are future and past null surfaces along with future timelike ( i + ) i + (i^(+))\left(i^{+}\right)(i+), past timelike ( i ) i (i^(-))\left(i^{-}\right)(i)and spacelike ( i 0 ) i 0 (i^(0))\left(i^{0}\right)(i0) infinities. The existence of these infinities in this case is due to the fact that the spacetime is asymptotically flat, and so we expect to recover something like the Minkowski conformal structure far at large distances from r = 0 r = 0 r=0r=0r=0. We also see how the singularity in the past is visible to observers in region I. This is known as a naked singularity, and can influence event in the future. Naked singularities are not a feature of normal physical processes and so its presence here is likely the result of the artificial extension we have made.
In Fig. 27.9(b), we have added to the Penrose diagram some lines of constant t t ttt (which run side to side in region I and up and down in region II) and lines of constant r r rrr (which run up and down in region I and side to side in region II). We can now justify the presence of trapped surfaces that we used to motivate the last chapter.

Example 27.5

In region I, a light pulse is emitted at point P P P\mathcal{P}P. At a later instant, defined by the future light cone (found by moving up to the next line running side to side), the outgoing wavefront s s sss has a larger radius (further to the right in the diagram). The ingoing wavefront q q qqq has a smaller radius (further to the left). This contrasts with the situation in region II. Here a light pulse is emitted at point Q Q Q\mathcal{Q}Q. At a later instant (again defined by the future light cone: move up to the next line running side to side) both outgoing s s sss and incoming q q qqq wavefronts are at a smaller value of r r rrr. This is because the next line up that runs side to side is a line of constant r r rrr and is smaller, since r r rrr decreases towards r = 0 r = 0 r=0r=0r=0, which is the singularity at the top of the diagram. We conclude that the areas of the spherical wavefronts decrease with time. These are trapped surfaces and warn us that there the wavefronts will meet a singularity in the future, as indeed they shall at r = 0 r = 0 r=0r=0r=0.
After two rather geometry-heavy chapters discussing the spacetime near a Schwarzschild black hole, we turn to the physics of the thermodynamics of the hole in the next chapter.

Chapter summary

  • Kruskal-Szekeres coordinates successfully remove the singularity at r S r S r_(S)r_{\mathrm{S}}rS. They suggest that the coordinates can be extended to cover apparently unphysical regions of spacetime.
  • Kruskal coordinates and their link to Schwarzschild coordinates can be understood by analogy to the relation between Minkowski coordinates and accelerating Minkowski coordinates.
  • The conformal structure of the system reveals how a physical singularity can be thought of as living at the edge of spacetime.

Exercises

(27.1) Verify eqn 27.2 .
(27.2) Show that the Kruskal coordinates are related to the Schwarzschild ones via
T 2 X 2 = ( 1 r 2 M ) e r / 2 M , (27.13) tanh 1 ( T X ) = t 4 M . T 2 X 2 = 1 r 2 M e r / 2 M , (27.13) tanh 1 T X = t 4 M . {:[T^(2)-X^(2)=(1-(r)/(2M))e^(r//2M)","],[(27.13)tanh^(-1)((T)/(X))=(t)/(4M).]:}\begin{align*} T^{2}-X^{2} & =\left(1-\frac{r}{2 M}\right) \mathrm{e}^{r / 2 M}, \\ \tanh ^{-1}\left(\frac{T}{X}\right) & =\frac{t}{4 M} . \tag{27.13} \end{align*}T2X2=(1r2M)er/2M,(27.13)tanh1(TX)=t4M.
(27.3) An accelerating particle in Minkowski space with coordinates ( T , X ) ( T , X ) (T,X)(T, X)(T,X) follows a world line X 2 T 2 = X 2 T 2 = X^(2)-T^(2)=X^{2}-T^{2}=X2T2=
g 2 g 2 g^(-2)g^{-2}g2 with
T ( τ ) = 1 g sinh g τ (27.14) X ( τ ) = 1 g cosh g τ T ( τ ) = 1 g sinh g τ (27.14) X ( τ ) = 1 g cosh g τ {:[T(tau)=(1)/(g)*sinh g tau],[(27.14)X(tau)=(1)/(g)*cosh g tau]:}\begin{align*} & T(\tau)=\frac{1}{g} \cdot \sinh g \tau \\ & X(\tau)=\frac{1}{g} \cdot \cosh g \tau \tag{27.14} \end{align*}T(τ)=1gsinhgτ(27.14)X(τ)=1gcoshgτ
Show that accelerating observers follow world lines with constant x x xxx in Rindler spacetime, which is described by a line element d s 2 = x 2 d t 2 + d x 2 d s 2 = x 2 d t 2 + d x 2 ds^(2)=-x^(2)dt^(2)+dx^(2)\mathrm{d} s^{2}=-x^{2} \mathrm{~d} t^{2}+\mathrm{d} x^{2}ds2=x2 dt2+dx2.
(27.4) Consider Rindler spacetime again with the metric given in the last question. Compute directly the
components of the acceleration a = u u a = u u a=grad_(u)u\boldsymbol{a}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}a=uu for a particle held at constant x = x 0 x = x 0 x=x_(0)x=x_{0}x=x0
Hint: The relevant connection coefficients are given in the answers to Exercise 9.4.
(27.5) We saw in this chapter how light from the event horizon was exponentially redshifted. We also saw how Schwarzschild coordinates can be thought of as accelerating Minkowski coordinates. We can link these ideas in very general terms, as we demonstrate here.
An observer moves in an inertial frame with trajectory X = v T X = v T X=vTX=v TX=vT
(a) Show that the observer's world line can be parametrized using T ( τ ) = γ τ T ( τ ) = γ τ T(tau)=gamma tauT(\tau)=\gamma \tauT(τ)=γτ and X ( τ ) = γ v τ X ( τ ) = γ v τ X(tau)=gamma v tauX(\tau)=\gamma v \tauX(τ)=γvτ. Now consider a light wave propagating with amplitude A = e i Ω ( T X ) A = e i Ω ( T X ) A=e^(iOmega(T-X))A=\mathrm{e}^{\mathrm{i} \Omega(T-X)}A=eiΩ(TX), with angular frequency Ω Ω Omega\OmegaΩ. (b) By using the parametrization in terms of τ τ tau\tauτ, show that the frequency measured by the observer
will be
(27.15) Ω = Ω ( 1 v 1 + v ) 1 2 . (27.15) Ω = Ω 1 v 1 + v 1 2 . {:(27.15)Omega^(')=Omega((1-v)/(1+v))^((1)/(2)).:}\begin{equation*} \Omega^{\prime}=\Omega\left(\frac{1-v}{1+v}\right)^{\frac{1}{2}} . \tag{27.15} \end{equation*}(27.15)Ω=Ω(1v1+v)12.
This is just the expected result for relativistic Doppler shift.
(c) Now consider a uniformly accelerated observer with
(27.16) T ( τ ) = 1 g sinh g τ , X ( τ ) = 1 g cosh g τ . (27.16) T ( τ ) = 1 g sinh g τ , X ( τ ) = 1 g cosh g τ . {:(27.16)T(tau)=(1)/(g)sinh g tau","quad X(tau)=(1)/(g)cosh g tau.:}\begin{equation*} T(\tau)=\frac{1}{g} \sinh g \tau, \quad X(\tau)=\frac{1}{g} \cosh g \tau . \tag{27.16} \end{equation*}(27.16)T(τ)=1gsinhgτ,X(τ)=1gcoshgτ.
Using the same procedure as above, show that the observer now measures the frequency of the wave as being exponentially redshifted according to
(27.17) ω ( τ ) = Ω e g τ (27.17) ω ( τ ) = Ω e g τ {:(27.17)omega(tau)=Omegae^(-g tau):}\begin{equation*} \omega(\tau)=\Omega \mathrm{e}^{-g \tau} \tag{27.17} \end{equation*}(27.17)ω(τ)=Ωegτ
We conclude the frequency decreases exponentially with proper time for the accelerated observer.

Hawking radiation

All nature wears one universal grin
Henry Fielding (1707-1754)

In 1974, Stephen Hawking 1 1 ^(1){ }^{1}1 showed that black holes are not simply holes in the Universe that consume matter. In fact, they continuously emit radiation just as a thermodynamic black body does. This effect is properly explained using quantum field theory: in particular, the notion that the ground state of a quantum field, known as the vacuum due to the absence of particles, allows vacuum fluctuations. The idea here is that it is possible to produce particle-antiparticle pairs that exist fleetingly before annihilating. The energy borrowed from the vacuum to produce the particles Δ E Δ E Delta E\Delta EΔE and the lifetime of the pair Δ t Δ t Delta t\Delta tΔt are constrained by an uncertainty relation Δ E Δ t Δ E Δ t Delta E Delta t~~ℏ\Delta E \Delta t \approx \hbarΔEΔt. The reason for this is that these virtual particles exist off mass shell, which is to say that for a particle with momentum p p p\boldsymbol{p}p and mass m m mmm, we no longer have the invariant relation p p = m 2 p p = m 2 p*p=-m^(2)\boldsymbol{p} \cdot \boldsymbol{p}=-m^{2}pp=m2. As a result, the energy of a particle is not determined by its usual dispersion relationship. Quantum uncertainty sets a time limit Δ t Δ E / Δ t Δ E / Delta t~~Delta E//ℏ\Delta t \approx \Delta E / \hbarΔtΔE/ on how long this can be accommodated, leading to the finite lifetime of the virtual particle and antiparticle pair.

28.1 Hawking radiation

The mechanism for black holes to radiate relies on a particle-antiparticle pair being produced just outside the event horizon of the hole, as shown in Fig. 28.1. One particle falls into the hole and is destroyed; the other propagates freely out to infinity and therefore constitutes the radiation. The idea is that if one particle from the pair crosses the event horizon to the region r < 2 M r < 2 M r < 2Mr<2 Mr<2M then the status of the pair can change. They were previously virtual particles forced to recombine within a time Δ t Δ t Delta t~~\Delta t \approxΔt / E / E ℏ//E\hbar / E/E. However, within the event horizon the changing roles of time and space can allow the (previously virtual) in-falling particle to propagate freely, no longer subject to the constraint on its lifetime. This also frees the outgoing particle. The resulting stream of real, outgoing particles from such processes transport energy from the hole and form Hawking radiation. This radiation has yet to be measured experimentally.
The key to understanding why this can occur is to examine the energymomentum of the virtual particle. For a particle to propagate with a momentum vector p p p\boldsymbol{p}p the observed energy E obs = p u E obs  = p u E_("obs ")=-p*uE_{\text {obs }}=-\boldsymbol{p} \cdot \boldsymbol{u}Eobs =pu, as measured by
28.1 Hawking radiation 289 28.2 Black-hole thermodynamics 292 Chapter summary 295 Exercises 296
1 1 ^(1){ }^{1}1 Stephen Hawking's advanced-level book The Large Scale Structure of Space-Time, cowritten with George F. R. Ellis (1939-), is warmly recommended for its clarity, insight, and ommended for its clarity, insight, and beautiful diagrams. Hawking is now
celebrated to the extent that he was celebrated to the extent that he was
voted the 25th greatest-ever Briton in voted the 25th greatest-ever Briton in
a poll conducted by the BBC in 2002, a poll conducted by the BBC in 2002 ,
one place below Queen Elizabeth II. one place below Queen Elizabeth 11 .
Of the other scientists mentioned in this book, Newton polled 6th, Faraday 22nd and Maxwell 91st. The poll was noted for several surprising results, including the placing of the actor Michael Crawford at seventeen, one place above Queen Victoria.
Fig. 28.1 Particle-antiparticle fluctuations near the event horizon of a black hole.
2 2 ^(2){ }^{2}2 It is often claimed that antiparticles have negative energy. In fact, all particles and antiparticles that have been observed have positive energy. The confusion arises as the antiparticles come from considering the two solutions to the dispersion relation equation E = ± ( p 2 + m 2 ) 1 2 E = ± p 2 + m 2 1 2 E=+-( vec(p)^(2)+m^(2))^((1)/(2))E= \pm\left(\vec{p}^{2}+m^{2}\right)^{\frac{1}{2}}E=±(p2+m2)12, which seems to allow negative energy states. In order to make physical sense, we must reinterpret the negative antiparticle solution with four-momentum p μ = ( E , p ) p μ = ( E , p ) p^(mu)=(-E, vec(p))p^{\mu}=(-E, \vec{p})pμ=(E,p). This, which describes a negative energy particle with outgoing momentum, is reinterpreted as a positive energy particle with incoming momentum p μ = p μ = p^(mu)=p^{\mu}=pμ= ( E , p E , p E,- vec(p)E,-\vec{p}E,p ), that is, we switch the sign of the momentum, so the antiparticle's direction is reversed, in order to guarantee positive energy.
3 3 ^(3){ }^{3}3 The link between the sign of energy and the passage of time can be seen in quantum mechanics by noting that quantum particles propagate with their wavefunction carrying a phase e i E t / e i E t / e^(-iEt//ℏ)\mathrm{e}^{-\mathrm{i} E t / \hbar}eiEt/ From the point of view of the wavefunc tion, a positive-energy particle propagating forward in time is equivalent to a negative energy one propagating backward in time.
4 4 ^(4){ }^{4}4 Such a path is timelike owing to the change in status of time and space inside the event horizon.
5 5 ^(5){ }^{5}5 Recall the effective energy equation for a photon is
1 b 2 = 1 L ~ 2 ( d r d λ ) 2 + 1 r 2 ( 1 2 M r ) 1 b 2 = 1 L ~ 2 d r d λ 2 + 1 r 2 1 2 M r (1)/(b^(2))=(1)/( tilde(L)^(2))*(((d)r)/((d)lambda))^(2)+(1)/(r^(2))(1-(2M)/(r))\frac{1}{b^{2}}=\frac{1}{\tilde{L}^{2}} \cdot\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}+\frac{1}{r^{2}}\left(1-\frac{2 M}{r}\right)1b2=1L~2( dr dλ)2+1r2(12Mr),
with b = L ~ / E ~ b = L ~ / E ~ b= tilde(L)// tilde(E)b=\tilde{L} / \tilde{E}b=L~/E~. For L ~ 0 L ~ 0 tilde(L)rarr0\tilde{L} \rightarrow 0L~0, we have that
E ~ = ± ( d r d λ ) = ± p r E ~ = ± d r d λ = ± p r tilde(E)=+-((dr)/((d)lambda))=+-p^(r)\tilde{E}= \pm\left(\frac{\mathrm{d} r}{\mathrm{~d} \lambda}\right)= \pm p^{r}E~=±(dr dλ)=±pr
6 6 ^(6){ }^{6}6 The observer has velocity with components u μ = ( 1 , 0 , 0 , 0 ) u μ = ( 1 , 0 , 0 , 0 ) u^(mu)=(1,0,0,0)u^{\mu}=(1,0,0,0)uμ=(1,0,0,0); the Killing vector ξ ξ xi\boldsymbol{\xi}ξ has components ξ μ = ( 1 , 0 , 0 , 0 ) ξ μ = ( 1 , 0 , 0 , 0 ) xi^(mu)=(1,0,0,0)\xi^{\mu}=(1,0,0,0)ξμ=(1,0,0,0). Recall that E ~ = ξ u E ~ = ξ u tilde(E)=-xi*u\tilde{E}=-\boldsymbol{\xi} \cdot \boldsymbol{u}E~=ξu is the energy per unit mass (hence a tilde on the E E EEE ). Therefore, E = ξ p E = ξ p E=-xi*pE=-\boldsymbol{\xi} \cdot \boldsymbol{p}E=ξp gives us the energy (no tilde) since p = m u p = m u p=mu\boldsymbol{p}=m \boldsymbol{u}p=mu, where m m mmm is the rest mass.
7 7 ^(7){ }^{7}7 That is, ξ ξ = e t e t = g t t = ( 1 ξ ξ = e t e t = g t t = ( 1 xi*xi=e_(t)*e_(t)=g_(tt)=-(1-\boldsymbol{\xi} \cdot \boldsymbol{\xi}=\boldsymbol{e}_{t} \cdot \boldsymbol{e}_{t}=g_{t t}=-(1-ξξ=etet=gtt=(1 2 M / r ) 2 M / r ) 2M//r)2 M / r)2M/r) changes sign on either side of the horizon at r = 2 M r = 2 M r=2Mr=2 Mr=2M.
8 8 ^(8){ }^{8}8 Of course, the observer is at infinity where they make measurements locally, and so they are not measuring this latter quantity.
a local observer with velocity u u u\boldsymbol{u}u, must be positive. 2 2 ^(2){ }^{2}2 If we do not make this stipulation, particles could travel backwards in time along timelike geodesics, which could allow violations of causality. 3 3 ^(3){ }^{3}3 The positivity of energy will be important in demonstrating the possibility of propagation inside the event horizon.

Example 28.1

We start by considering photons. A photon is identical to an antiphoton, so we can consider one member of a pair of virtual photons, with momentum p p p\boldsymbol{p}p, produced by a vacuum fluctuation just outside r = 2 M r = 2 M r=2Mr=2 Mr=2M. We will show how it can propagate freely at radii r < 2 M r < 2 M r < 2Mr<2 Mr<2M.
To do this, we put ourselves in the (unenviable) position of an observer inside the horizon, undergoing a radial plunge that follows a curve with tangent vector u u u\boldsymbol{u}u. To make things as simple as possible we choose a plunging path with u 0 = 0 u 0 = 0 u^(0)=0u^{0}=0u0=0, so that the only non-zero component of u u u\boldsymbol{u}u is 4 u r 4 u r ^(4)u^(r){ }^{4} u^{r}4ur. From the identity u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1, we have, for r < 2 M r < 2 M r < 2Mr<2 Mr<2M, that
(28.1) u r = ( 2 M r 1 ) 1 2 (28.1) u r = 2 M r 1 1 2 {:(28.1)u^(r)=-((2M)/(r)-1)^((1)/(2)):}\begin{equation*} u^{r}=-\left(\frac{2 M}{r}-1\right)^{\frac{1}{2}} \tag{28.1} \end{equation*}(28.1)ur=(2Mr1)12
The condition for a photon trajectory to be allowed is that the photon has a positive energy, as measured by us locally. This means that we require that its energy obeys E obs = p u > 0 E obs = p u > 0 E_(obs)=-p*u > 0E_{\mathrm{obs}}=-\boldsymbol{p} \cdot \boldsymbol{u}>0Eobs=pu>0. Let's consider a photon that, like us, is on a L = 0 L = 0 L=0L=0L=0 radial plunge and has 5 E ~ = ± p r 5 E ~ = ± p r ^(5) tilde(E)=+-p^(r){ }^{5} \tilde{E}= \pm p^{r}5E~=±pr. Its energy, measured by us, is then
(28.2) E obs = p u = g r r p r u r = ( 2 M r 1 ) 1 2 p r . (28.2) E obs = p u = g r r p r u r = 2 M r 1 1 2 p r . {:(28.2)E_(obs)=-p*u=-g_(rr)p^(r)u^(r)=-((2M)/(r)*-1)^(-(1)/(2))p^(r).:}\begin{equation*} E_{\mathrm{obs}}=-\boldsymbol{p} \cdot \boldsymbol{u}=-g_{r r} p^{r} u^{r}=-\left(\frac{2 M}{r} \cdot-1\right)^{-\frac{1}{2}} p^{r} . \tag{28.2} \end{equation*}(28.2)Eobs=pu=grrprur=(2Mr1)12pr.
This is only positive if p r < 0 p r < 0 p^(r) < 0p^{r}<0pr<0, which is always true for an in-falling photon. The key here is that this puts no constraint on E = p t E = p t E=p^(t)E=p^{t}E=pt, the timelike component of the photon momentum in flat spacetime. This effectively frees the photon from the constraints of the uncertainty relation: it can propagate freely inside the event horizon. The other member of the particle-antiparticle pair, an outgoing photon, is also then free to propagate.
We conclude that virtual photons can propagate inside the event horizon, effectively removing the constraint that the particle-antiparticle pair has a limited lifespan.
In order to understand how this leads to outgoing radiation, and hence energy being transported out of the hole, we now consider the fate of a massive particle and antiparticle pair, and the energy measured by an observer at rest at infinity. Since this observer's velocity 6 6 ^(6){ }^{6}6 is identical to the Killing vector ξ ξ xi\boldsymbol{\xi}ξ, the observer measures the energy E obs = p ξ E obs = p ξ E_(obs)^(oo)=-p*xiE_{\mathrm{obs}}^{\infty}=-\boldsymbol{p} \cdot \boldsymbol{\xi}Eobs=pξ, which is a quantity conserved along the particle's geodesic. However, the status of the Killing vector ξ = e t ξ = e t xi=e_(t)\boldsymbol{\xi}=\boldsymbol{e}_{\boldsymbol{t}}ξ=et changes from timelike outside the horizon to spacelike within it. 7 7 ^(7){ }^{7}7 This allows the possibility that the quantity p ξ p ξ -p*xi-\boldsymbol{p} \cdot \boldsymbol{\xi}pξ is negative inside the event horizon without it being unphysical. 8 8 ^(8){ }^{8}8 Therefore, if a particle-antiparticle pair is produced close to the event horizon, to ensure the positivity of p ξ p ξ -p*xi-\boldsymbol{p} \cdot \boldsymbol{\xi}pξ for the particle measured by the observer at infinity, we require that the particle with negative p ξ p ξ -p*xi-\boldsymbol{p} \cdot \boldsymbol{\xi}pξ falls into the hole, while one with positive p ξ p ξ -p*xi-\boldsymbol{p} \cdot \boldsymbol{\xi}pξ (of equal magnitude) travels out to infinity, where it is measured, locally, by the observer. This guarantees that the observer at infinity measures
a positive energy particle and that, for the observer at infinity, energy has been conserved. This might seem to contradict the idea that we do not interpret any particles/antiparticles in Nature as carrying negative energy. However, it is really a symptom of the geometrical property of the quantity p ξ p ξ -p*xi-\boldsymbol{p} \cdot \boldsymbol{\xi}pξ, unique to black holes, rather than an admission of negative energy particles as measured locally. As we saw in the last example, measured locally, the energy of propagating particles and antiparticles is necessarily positive.
This curious state of affairs allows the outgoing particle produced in the pair to propagate outside the horizon and to carry its energy out to infinity. We compute how much energy in the next example.

Example 28.2

We, the observers, will watch a single vacuum fluctuation that creates a photon pair just outside the horizon. We work in a freely falling reference frame that is momentarily at rest at r = 2 M + ε r = 2 M + ε r=2M+epsir=2 M+\varepsilonr=2M+ε, that subsequently falls inwards following a trajectory with L ~ = 0 L ~ = 0 tilde(L)=0\tilde{L}=0L~=0 and 9 9 ^(9)^{9}9
(28.4) E ~ = ( 1 2 M 2 M + ε ) 1 2 ( ε 2 M ) 1 2 (28.4) E ~ = 1 2 M 2 M + ε 1 2 ε 2 M 1 2 {:(28.4) tilde(E)=(1-(2M)/(2M+epsi))^((1)/(2))~~((epsi)/(2M))^((1)/(2)):}\begin{equation*} \tilde{E}=\left(1-\frac{2 M}{2 M+\varepsilon}\right)^{\frac{1}{2}} \approx\left(\frac{\varepsilon}{2 M}\right)^{\frac{1}{2}} \tag{28.4} \end{equation*}(28.4)E~=(12M2M+ε)12(ε2M)12
The proper time for the frame to fall to the horizon is given (rearranging eqn 28.3) by
(28.5) Δ τ = 2 M + ε 2 M d r ( 2 M r 2 M 2 M + ε ) 1 2 2 ( 2 M ε ) 1 2 (28.5) Δ τ = 2 M + ε 2 M d r 2 M r 2 M 2 M + ε 1 2 2 ( 2 M ε ) 1 2 {:(28.5)Delta tau=-int_(2M+epsi)^(2M)((d)r)/(((2M)/(r)-(2M)/(2M+epsi))^((1)/(2)))~~2(2M epsi)^((1)/(2)):}\begin{equation*} \Delta \tau=-\int_{2 M+\varepsilon}^{2 M} \frac{\mathrm{~d} r}{\left(\frac{2 M}{r}-\frac{2 M}{2 M+\varepsilon}\right)^{\frac{1}{2}}} \approx 2(2 M \varepsilon)^{\frac{1}{2}} \tag{28.5} \end{equation*}(28.5)Δτ=2M+ε2M dr(2Mr2M2M+ε)122(2Mε)12
The photon must, at this stage of the process, exist subject to the Heisenberg uncertainty relation Δ E Δ τ Δ E Δ τ Delta E Delta tau~~ℏ\Delta E \Delta \tau \approx \hbarΔEΔτ, which means it has an energy, as measured in our local inertial frame, of
(28.6) E obs 2 ( 2 M ε ) 1 2 (28.6) E obs 2 ( 2 M ε ) 1 2 {:(28.6)E_(obs)~~(ℏ)/(2(2M epsi)^((1)/(2))):}\begin{equation*} E_{\mathrm{obs}} \approx \frac{\hbar}{2(2 M \varepsilon)^{\frac{1}{2}}} \tag{28.6} \end{equation*}(28.6)Eobs2(2Mε)12
Once the in-falling particle has crossed the horizon, the outgoing photon disappears to infinity taking its energy with it. Let's find this energy at infinity. The conserved quantity on the photon's trajectory is p ξ = p t p ξ = p t p*xi=p_(t)\boldsymbol{p} \cdot \boldsymbol{\xi}=p_{t}pξ=pt, which is also minus the energy E obs E obs  E_("obs ")^(oo)E_{\text {obs }}^{\infty}Eobs , measured by the observer at infinity. This can be linked with the energy measured by us locally in the frame just outside the horizon. We have that the energy in the local frame E obs = p u E obs = p u E_(obs)=-p*uE_{\mathrm{obs}}=-\boldsymbol{p} \cdot \boldsymbol{u}Eobs=pu with the only non-zero component of our instantaneous velocity 10 u t = E ~ ( ε / 2 M ) 1 2 10 u t = E ~ ( ε / 2 M ) 1 2 ^(10)-u_(t)= tilde(E)~~(epsi//2M)^((1)/(2)){ }^{10}-u_{t}=\tilde{E} \approx(\varepsilon / 2 M)^{\frac{1}{2}}10ut=E~(ε/2M)12. We therefore have
(28.7) E obs = g t t p t u t = g t t E obs E ~ (28.7) E obs = g t t p t u t = g t t E obs E ~ {:(28.7)E_(obs)=-g^(tt)p_(t)u_(t)=-g^(tt)E_(obs)^(oo) tilde(E):}\begin{equation*} E_{\mathrm{obs}}=-g^{t t} p_{t} u_{t}=-g^{t t} E_{\mathrm{obs}}^{\infty} \tilde{E} \tag{28.7} \end{equation*}(28.7)Eobs=gttptut=gttEobsE~
where g t t g t t g^(tt)g^{t t}gtt is evaluated at r = 2 M + ε r = 2 M + ε r=2M+epsir=2 M+\varepsilonr=2M+ε. For a Schwarzschild black hole
(28.8) g t t = [ 1 2 M 2 M + ε ] 1 = ( 1 + 2 M ε ) 2 M ε (28.8) g t t = 1 2 M 2 M + ε 1 = 1 + 2 M ε 2 M ε {:(28.8)g^(tt)=-[1-(2M)/(2M+epsi)]^(-1)=-(1+(2M)/(epsi))~~-(2M)/(epsi):}\begin{equation*} g^{t t}=-\left[1-\frac{2 M}{2 M+\varepsilon}\right]^{-1}=-\left(1+\frac{2 M}{\varepsilon}\right) \approx-\frac{2 M}{\varepsilon} \tag{28.8} \end{equation*}(28.8)gtt=[12M2M+ε]1=(1+2Mε)2Mε
This gives us, plugging into eqn 28.7,
(28.9) 2 ( 2 M ϵ ) 1 2 = ( 2 M ε ) E obs ( ε 2 M ) 1 2 (28.9) 2 ( 2 M ϵ ) 1 2 = 2 M ε E obs ε 2 M 1 2 {:(28.9)(ℏ)/(2(2M epsilon)^((1)/(2)))=((2M)/(epsi))E_(obs)^(oo)((epsi)/(2M))^((1)/(2)):}\begin{equation*} \frac{\hbar}{2(2 M \epsilon)^{\frac{1}{2}}}=\left(\frac{2 M}{\varepsilon}\right) E_{\mathrm{obs}}^{\infty}\left(\frac{\varepsilon}{2 M}\right)^{\frac{1}{2}} \tag{28.9} \end{equation*}(28.9)2(2Mϵ)12=(2Mε)Eobs(ε2M)12
or
(28.10) E obs = 4 M (28.10) E obs = 4 M {:(28.10)E_(obs)^(oo)=(ℏ)/(4M):}\begin{equation*} E_{\mathrm{obs}}^{\infty}=\frac{\hbar}{4 M} \tag{28.10} \end{equation*}(28.10)Eobs=4M
In conclusion, the photon takes an amount of energy out to infinity which is inversely proportional to the mass of the black hole.
9 9 ^(9){ }^{9}9 Recall that
E ~ 2 1 2 = 1 2 ( u r ) 2 + V eff ( r ) , E ~ 2 1 2 = 1 2 u r 2 + V eff ( r ) , ( tilde(E)^(2)-1)/(2)=(1)/(2)(u^(r))^(2)+V_(eff)(r),\frac{\tilde{E}^{2}-1}{2}=\frac{1}{2}\left(u^{r}\right)^{2}+V_{\mathrm{eff}}(r),E~212=12(ur)2+Veff(r),
and so in this case, where L ~ = 0 L ~ = 0 tilde(L)=0\tilde{L}=0L~=0, we have
(28.3) ( d r d τ ) 2 = E ~ 2 1 + 2 M r (28.3) d r d τ 2 = E ~ 2 1 + 2 M r {:(28.3)((dr)/((d)tau))^(2)= tilde(E)^(2)-1+(2M)/(r):}\begin{equation*} \left(\frac{\mathrm{d} r}{\mathrm{~d} \tau}\right)^{2}=\tilde{E}^{2}-1+\frac{2 M}{r} \tag{28.3} \end{equation*}(28.3)(dr dτ)2=E~21+2Mr
If u r = 0 u r = 0 u^(r)=0u^{r}=0ur=0 at r = 2 M + ε r = 2 M + ε r=2M+epsir=2 M+\varepsilonr=2M+ε we obtain the quoted result.
10 10 ^(10){ }^{10}10 Recall that
E ~ = ξ u = u t E ~ = ξ u = u t tilde(E)=-xi*u=-u_(t)\tilde{E}=-\boldsymbol{\xi} \cdot \boldsymbol{u}=-u_{t}E~=ξu=ut
11 11 ^(11){ }^{11}11 The area A A AAA can be computed from the Schwarzschild metric (see exercises).
12 12 ^(12){ }^{12}12 We have assumed here that the black hole is not consuming energy from the outside Universe as it emits Hawkin outside Universe as it emits Hawking radiation. However, our Universe is aglow with the left-over radiation from
the Big Bang. The cosmic microwave the Big Bang. The cosmic microwave background (CMB, Chapter 15) with a black-body temperature of T CMB = T CMB = T_(CMB)=T_{\mathrm{CMB}}=TCMB= 2.73 K , and this radiation continually feeds any black hole. (This is in addition to the effect of nearby dust or gas, which dominates the growth of black holes that have been observed in the Universe.) Black holes with temperatures T < T CMB T < T CMB T < T_(CMB)T<T_{\mathrm{CMB}}T<TCMB therefore cannot actually shrink due to Hawking ranot actually shrink due to Hawking ra-
diation, even in principle, but must diation, even in principle, but must
continue to grow, at least until that continue to grow, at least until that
point in time when the CMB and other point in time when the CMB and other
sources of cosmic radiation sufficiently sources of cosmic radiation sufficiently
cool down to temperatures below that of the black hole. This is compatible with the second law of thermodynamics, which tells us that heat flows from a hotter object to a colder one, and not the other way round To get a sense of how small a black hole would sense of how small a black hole would have to be such that its temperature would exceed that of the present-day CMB, we can use eqn 28.12 to calculate the Hawking temperature of black hole with one solar mass, M = 1.99 × M = 1.99 × M_(o.)=1.99 xxM_{\odot}=1.99 \timesM=1.99× 10 30 kg 10 30 kg 10^(30)kg10^{30} \mathrm{~kg}1030 kg. This yields a temperature of just 62 nK T CMB 62 nK T CMB 62nK≪T_(CMB)62 \mathrm{nK} \ll T_{\mathrm{CMB}}62nKTCMB. In fact, using eqn 28.12 to calculate the upper bound on the black hole's mass such that its temperature would at least match that of the CMB, i.e. setting T = T CMB T = T CMB T=T_(CMB)T=T_{\mathrm{CMB}}T=TCMB, we obtain M CMB = 4.5 × 10 23 kg = M CMB = 4.5 × 10 23 kg = M_(CMB)=4.5 xx10^(23)kg=M_{\mathrm{CMB}}=4.5 \times 10^{23} \mathrm{~kg}=MCMB=4.5×1023 kg= 2.3 × 10 8 M 2.3 × 10 8 M 2.3 xx10^(-8)M_(o.)2.3 \times 10^{-8} M_{\odot}2.3×108M. Any black hole with a mass larger than M CMB M CMB  M_("CMB ")M_{\text {CMB }}MCMB  will thus continue to grow for quite some time yet at least in the present-day universe.
13 13 ^(13){ }^{13}13 A Killing horizon is a sort of generalized geometrical upgrade of the idea of an event horizon. It is a null hypersurface defined by the vanishing of the norm of a Killing vector field ξ ξ xi\boldsymbol{\xi}ξ. This bit of formalism allows a physical definition of the surface gravity κ κ kappa\kappaκ of a Killing horizon: it is the acceleration, exerted at infinity, needed to keep an object at the Killing horizon. See Poisson for proof of this and justification of eqn 28.13. The logic here is to choose ξ ξ xi\boldsymbol{\xi}ξ to give a suitable Killing horizon on which to evaluate the surface gravity.
Although we have been dealing with a vastly simplified picture, the rigorous quantum field theory computation gives a result that is (perhaps surprisingly) fairly close to the estimate from the previous example. We examine the consequences of this in the next section.

28.2 Black-hole thermodynamics

The rigorous result is that the typical outgoing photon energy measured at infinity is actually E obs = / 8 π M E obs = / 8 π M E_(obs)^(oo)=ℏ//8pi ME_{\mathrm{obs}}^{\infty}=\hbar / 8 \pi MEobs=/8πM, but that they can be detected with exactly the range of energies that one would expect for a thermodynamic black body. As a result, black holes are black bodies with a characteristic temperature
(28.11) k B T 8 π M . (28.11) k B T 8 π M . {:(28.11)k_(B)T~~(ℏ)/(8pi M).:}\begin{equation*} k_{\mathrm{B}} T \approx \frac{\hbar}{8 \pi M} . \tag{28.11} \end{equation*}(28.11)kBT8πM.
Since the hole is emitting radiation, it must be losing energy and consequently mass. This is consistent with the negativity of p ξ p ξ -p*xi-\boldsymbol{p} \cdot \boldsymbol{\xi}pξ for the member of the particle-antiparticle pair that falls into the hole. Let's look into this.

Example 28.3

The temperature of a black hole is proportional to M 1 M 1 M^(-1)M^{-1}M1. For any black body, the rate of emission of energies varies as A T 4 A T 4 AT^(4)A T^{4}AT4, where A A AAA is the area of the emitter, in this case the area of the black hole's event horizon. 11 11 ^(11){ }^{11}11 For the Schwarzschild black hole, the area is A = 16 π M 2 A = 16 π M 2 A=16 piM^(2)A=16 \pi M^{2}A=16πM2. The luminosity therefore goes as M 2 M 2 M^(-2)M^{-2}M2, and is the result of a decrease in the mass of the hole, and so we can write
(28.12) d M d t 1 M 2 (28.12) d M d t 1 M 2 {:(28.12)(dM)/((d)t)prop-(1)/(M^(2)):}\begin{equation*} \frac{\mathrm{d} M}{\mathrm{~d} t} \propto-\frac{1}{M^{2}} \tag{28.12} \end{equation*}(28.12)dM dt1M2
Solving this we see that the decrease in mass of the hole with time leads us to conclude that the lifetime τ τ tau\tauτ of a black hole must vary as 12 τ M 3 12 τ M 3 ^(12)tau∼M^(3){ }^{12} \tau \sim M^{3}12τM3.
The last example demonstrates that black-hole radiation causes an isolated black hole to effectively evaporate, with its mass decreasing and temperature increasing until, presumably, it disappears. We return to the consequence of this later.
We have the result that, through Hawking's proposed mechanism, a black hole radiates such that it is a black body with a temperature T T TTT. In fact, this is only the start of a complete set of thermodynamic laws that govern the mechanics of black holes. These follows from the fact that the temperature T T TTT is equivalent to another property of the black hole: the surface gravity κ κ kappa\kappaκ. This is defined geometrically as the quantity
(28.13) ξ μ ξ ; μ ν = κ ξ ν (28.13) ξ μ ξ ; μ ν = κ ξ ν {:(28.13)xi^(mu)xi_(;mu)^(nu)=kappaxi^(nu):}\begin{equation*} \xi^{\mu} \xi_{; \mu}^{\nu}=\kappa \xi^{\nu} \tag{28.13} \end{equation*}(28.13)ξμξ;μν=κξν
evaluated at the event horizon. Here ξ ξ xi\boldsymbol{\xi}ξ is a suitably chosen linear combination of Killing vectors. 13 13 ^(13){ }^{13}13

Example 28.4

For the Schwarzschild solution we have a Killing vector in Schwarzschild coordinates ( t , r , θ , ϕ ) ( t , r , θ , ϕ ) (t,r,theta,phi)(t, r, \theta, \phi)(t,r,θ,ϕ) of ξ μ = ( 1 , 0 , 0 , 0 ) ξ μ = ( 1 , 0 , 0 , 0 ) xi^(mu)=(1,0,0,0)\xi^{\mu}=(1,0,0,0)ξμ=(1,0,0,0). Using the incoming Eddington-Finkelstein coordinates ( V , r , θ , ϕ V , r , θ , ϕ V,r,theta,phiV, r, \theta, \phiV,r,θ,ϕ ) examined in Chapter 26, we find that this quantity has components ξ ν = ξ ν = xi^(nu)=\xi^{\nu}=ξν= ( 1 , 0 , 0 , 0 ) . 14 ( 1 , 0 , 0 , 0 ) . 14 (1,0,0,0).^(14)(1,0,0,0) .{ }^{14}(1,0,0,0).14 On carrying out the necessary computation using eqn 28.13 , we obtain
(28.15) r ( 1 + 2 M r ) M r 2 = κ (28.15) r 1 + 2 M r M r 2 = κ {:(28.15)-(del)/(del r)(-1+(2M)/(r))-(M)/(r^(2))=kappa:}\begin{equation*} -\frac{\partial}{\partial r}\left(-1+\frac{2 M}{r}\right)-\frac{M}{r^{2}}=\kappa \tag{28.15} \end{equation*}(28.15)r(1+2Mr)Mr2=κ
giving, on the surface r = 2 M r = 2 M r=2Mr=2 Mr=2M, a surface gravity
(28.16) κ = 1 4 M . (28.16) κ = 1 4 M . {:(28.16)kappa=(1)/(4M).:}\begin{equation*} \kappa=\frac{1}{4 M} . \tag{28.16} \end{equation*}(28.16)κ=14M.
Since the temperature is given by k B T = / 8 π M k B T = / 8 π M k_(B)T=ℏ//8pi Mk_{\mathrm{B}} T=\hbar / 8 \pi MkBT=/8πM, we can relate the temperature and
the surface gravity by
k B T = κ 2 π k B T = κ 2 π k_(B)T=(ℏkappa)/(2pi)k_{\mathrm{B}} T=\frac{\hbar \kappa}{2 \pi}kBT=κ2π
So surface gravity and temperature are proportional.
The last example shows that surface gravity and temperature are effectively equivalent for a black hole.
In classical thermodynamics, 15 15 ^(15){ }^{15}15 the conjugate variable of temperature is entropy S S SSS. The property of the black hole analogous to the entropy is the area A A AAA of the event horizon. To see this we start with the first law of thermodynamics, which says that, in the absence of any work being done, we have d U = T d S d U = T d S dU=TdS\mathrm{d} U=T \mathrm{~d} SdU=T dS. Taking the mass M M MMM of the hole as a measure of its internal energy, we have for a Schwarzschild black hole that the area A = 16 π M 2 A = 16 π M 2 A=16 piM^(2)A=16 \pi M^{2}A=16πM2 and so we can write
(28.18) d M = d A 32 π M = 8 π k B M d ( A k B 4 ) . (28.18) d M = d A 32 π M = 8 π k B M d A k B 4 . {:(28.18)dM=(dA)/(32 pi M)=(ℏ)/(8pik_(B)M)d((Ak_(B))/(4ℏ)).:}\begin{equation*} \mathrm{d} M=\frac{\mathrm{d} A}{32 \pi M}=\frac{\hbar}{8 \pi k_{\mathrm{B}} M} \mathrm{~d}\left(\frac{A k_{\mathrm{B}}}{4 \hbar}\right) . \tag{28.18} \end{equation*}(28.18)dM=dA32πM=8πkBM d(AkB4).
Since T = / 8 π k B M T = / 8 π k B M T=ℏ//8pik_(B)MT=\hbar / 8 \pi k_{\mathrm{B}} MT=/8πkBM is the temperature, we can use the resemblance to d U = T d S d U = T d S dU=TdS\mathrm{d} U=T \mathrm{~d} SdU=T dS to link entropy S S SSS and area A A AAA via
(28.19) S = A k B 4 . (28.19) S = A k B 4 . {:(28.19)S=(Ak_(B))/(4ℏ).:}\begin{equation*} S=\frac{A k_{\mathrm{B}}}{4 \hbar} . \tag{28.19} \end{equation*}(28.19)S=AkB4.
We now generalize by stating the laws of black hole mechanics. For a black hole of mass M M MMM, angular momentum L L LLL, charge q q qqq, horizon area A A AAA and surface gravity κ κ kappa\kappaκ, we have the four laws: 16 16 ^(16){ }^{16}16
Oth law The surface gravity κ κ kappa\kappaκ is constant over the horizon.
1st law For two stationary black holes that differ only by small variations in M , L M , L M,LM, LM,L and q q qqq, we have
(28.20) d M = κ 8 π d A + Ω d L + Φ d q , (28.20) d M = κ 8 π d A + Ω d L + Φ d q , {:(28.20)dM=(kappa)/(8pi)dA+OmegadL+Phidq",":}\begin{equation*} \mathrm{d} M=\frac{\kappa}{8 \pi} \mathrm{~d} A+\Omega \mathrm{d} L+\Phi \mathrm{d} q, \tag{28.20} \end{equation*}(28.20)dM=κ8π dA+ΩdL+Φdq,
where Ω Ω Omega\OmegaΩ and Φ Φ Phi\PhiΦ are, respectively, the angular velocity and electric potential at the horizon.
2nd law The area A A AAA of the horizon never decreases, which is to say that d A 0 d A 0 dA >= 0\mathrm{d} A \geq 0dA0, as a function of cosmic time.
3rd law It is impossible to reduce the surface gravity to zero in a finite number of steps.
14 14 ^(14){ }^{14}14 Since we need to take a derivative, it turns out that we need to compute the components of the corresponding 1 form to obtain
(28.14) ξ ν = ( 1 + 2 M r , 1 , 0 , 0 ) (28.14) ξ ν = 1 + 2 M r , 1 , 0 , 0 {:(28.14)xi_(nu)=(-1+(2M)/(r),1,0,0):}\begin{equation*} \xi_{\nu}=\left(-1+\frac{2 M}{r}, 1,0,0\right) \tag{28.14} \end{equation*}(28.14)ξν=(1+2Mr,1,0,0)
This computation is discussed in the Exercises. We can see that this Killing field is null on the event horizon, giving us the Killing horizon on which we evaluate the surface gravity
15 15 ^(15){ }^{15}15 See Chapter 39 for further discussion of thermodynamics.
16 16 ^(16){ }^{16}16 The laws of thermodynamics are: Oth law The temperature is the same in interacting systems at equilibrium, 1st law d U = T d S p d V d U = T d S p d V dU=TdS-pdV\mathrm{d} U=T \mathrm{~d} S-p \mathrm{~d} VdU=T dSp dV
2nd law d S ( t ) 0 d S ( t ) 0 dS(t) >= 0\mathrm{d} S(t) \geq 0dS(t)0
3rd law It is impossible to reduce T T TTT to zero in a finite number of steps.
17 17 ^(17){ }^{17}17 We have the correspondences with conventional thermodynamics (assuming a sign convention for the work term d W = p d V ) d W = p d V ) dW=-pdV)\mathrm{d} W=-p \mathrm{~d} V)dW=p dV) :
(28.21) M U , κ T , A S , Ω d L d W , = d W , (28.21) M U , κ T , A S , Ω d L d W , = d W , {:(28.21){:[M,-=,U","],[kappa,-=,T","],[A,-=,S","],[OmegadL,-=,-dW","],[,=,dW","]:}:}\begin{array}{clc} M & \equiv & U, \tag{28.21}\\ \kappa & \equiv & T, \\ A & \equiv & S, \\ \Omega \mathrm{~d} L & \equiv & -\mathrm{d} W, \\ & = & \mathrm{d} W, \end{array}(28.21)MU,κT,AS,Ω dLdW,=dW,
18 18 ^(18){ }^{18}18 For more on black hole thermodynamics, see the excellent review articles by Page (2005) and Carlip (2014) in the Further Reading in Appendix A.
19 19 ^(19){ }^{19}19 Hawking offered two quotations for inclusion on the cover of the spoof newspaper The Framley Examiner: 'This is absolute rubbish' and 'Laughed briefly'. The first was used.
Fig. 28.2 Penrose diagram showing spacetime featuring a completely evaporating black hole.
These four laws bear an intended resemblance to the four laws of thermodynamics. 17 17 ^(17){ }^{17}17 One quite dramatic consequence of the second law is that two colliding black holes must give rise to a hole that has an horizon area that is greater than or equal to the areas of the two holes we started with.
The second law demonstrates that when matter falls into a black hole, even though we lose the entropy of this matter (we no longer have access to its thermodynamic degrees of freedom), the increase in the black hole's mass M M MMM leads to an increase in area A A AAA and hence entropy S S SSS which means the entropy of the Universe has a net increase. We might wonder whether the second law of thermodynamics is still compatible with the decrease in mass, and hence area, that occurs when the black hole radiates; this process clearly violates d A 0 d A 0 dA >= 0\mathrm{d} A \geq 0dA0, the formulation of the second law we have given above. For this reason, we need to invoke a generalized second law which applies to the physical situation more holistically than just considering only the black hole. This new law states that the total entropy of a black hole and its environment cannot decrease. Once the emitted radiation from a black hole has thermalized to the temperature of the external heat bath (one can perhaps assume the rest of the Universe acts as some kind of thermal reservoir) then the law implies that the entropy resulting from the radiation should at least compensate the decrease in entropy of the shrinking black hole. This result can be proved rigorously, albeit by making certain assumptions. 18 18 ^(18){ }^{18}18
We might also wonder about the status of information. Since objects cannot be seen after passing into the event horizon of a black hole, the information about them seems to be lost. As the black hole evaporates, it seems that this loss might be permanent. This is a worry since, in quantum mechanics, wavefunctions evolve unitarily so that the state of the system at a later time can always be deduced from that at an earlier time, and vice versa. Irreversible destruction of information about a physical system appears to break this property. This has been deemed so potentially disturbing that it has led some to propose that the end point of an evaporating black hole might not be its disappearance. There have been proposals that the information is permanently stored in a nugget, left over at the end of the evaporation, or that the death of a black hole is accompanied by an explosion that returns the information to the Universe. Hawking's own opinion was that the loss was permanent. 19 19 ^(19){ }^{19}19

Example 28.5

As the black hole radiates its mass reduces and the hole effectively evaporates. Whether this can continue until the black hole ceases to exist is another open question. This is because, as the black hole becomes small, a classical theory such as general relativity should not be expected to describe the physics and so we expect to need to employ quantum gravity (Chapter 49), which does not yet exist as a complete theory. Nevertheless, if we accept that the black hole will vanish, spacetime will be described by the Penrose diagram shown in Fig. 28.2.
Before moving on, it's worth pausing to recall from Chapter 26 that Schwarzschild coordinates can be though of as being equivalent to accelerating Minkowski coordinates. This suggests that some of the subtle behaviour that occurs near a black hole might have an analogy in a system of accelerating particles. This is the basis of the Unruh effect 20 20 ^(20){ }^{20}20 which predicts that owing to quantum fluctuations in the vacuum, an observer accelerating through empty space with no other heat sources present will still observe black body radiation (i.e. the observer will think they are moving through a thermal bath), whereas an unaccelerated (inertial) observer will not observe any temperature. The Unruh temperature of this perceived blackbody radiation is given by a / ( 2 π c k B ) a / 2 π c k B ℏa//(2pi ck_(B))\hbar a /\left(2 \pi c k_{\mathrm{B}}\right)a/(2πckB), where a a aaa is the acceleration of the observer.

Example 28.6

Close to the event horizon of a Schwarzschild black hole, the spacetime can be expressed in terms of Rindler coordinates which has line element
(28.22) d s 2 = x 2 d t 2 + d x 2 (28.22) d s 2 = x 2 d t 2 + d x 2 {:(28.22)ds^(2)=-x^(2)dt^(2)+dx^(2):}\begin{equation*} \mathrm{d} s^{2}=-x^{2} \mathrm{~d} t^{2}+\mathrm{d} x^{2} \tag{28.22} \end{equation*}(28.22)ds2=x2 dt2+dx2
Accelerated observers in Minkowski spacetime 21 21 ^(21){ }^{21}21 have world lines that coincide with lines of constant x x xxx in the Rindler spacetime. As discussed in Chapter 26, and shown in Fig. 28.3, Rindler coordinates only cover that part of Minkowski spacetime with X > | T | X > | T | X > |T|X>|T|X>|T|. Nothing at X < | T | X < | T | X < |T|X<|T|X<|T| can ever catch up with the accelerated observer, so the line T = ± X T = ± X T=+-XT= \pm XT=±X is an effective event horizon for the accelerated observer. With this setup we can rerun the argument for Hawking radiation. A particle-antiparticle pair is created at the event horizon T = X T = X T=XT=XT=X. One of the pair passes through the horizon into the region X < T X < T X < TX<TX<T where it is lost to the observer, leaving the other free to escape into the allowed region X > | T | X > | T | X > |T|X>|T|X>|T|. This free particle is then responsible for the Unruh radiation. The temperature of the radiation for an observer with acceleration a a aaa can be estimated via dimensional analysis to be k B T a / c k B T a / c k_(B)T∼ℏa//ck_{\mathrm{B}} T \sim \hbar a / ckBTa/c.

Chapter summary

  • Black holes radiate Hawking radiation through a process where a particle-antiparticle pair is created at the event horizon. Energy is transported out to infinity.
  • Black holes are thermodynamic black bodies with temperatures T M 1 T M 1 T propM^(-1)T \propto M^{-1}TM1, where M M MMM is the mass of the hole.
  • A set of laws of black hole mechanics can be formulated that closely resemble the laws of thermodynamics. Surface gravity is analogous to temperature and the area of the event horizon is analogous to entropy.
    20 20 ^(20){ }^{20}20 W. G. 'Bill' Unruh (1945- ). Speaking about his effect, Unruh quipped that it shows that you could cook your steak just by accelerating it. He also admitted that an acceleration of 10 24 cm s 1 10 24 cm s 1 10^(24)cms^(-1)10^{24} \mathrm{~cm} \mathrm{~s}^{-1}1024 cm s1 would be required to achieve a temperature of 300 C 300 C 300^(@)C300^{\circ} \mathrm{C}300C, which one could also obtain from a good charcoal grill.
Fig. 28.3 An accelerated observer follows a hyperbolic path in Minkowski spacetime. At elevated values of acceleration they are close to a horizon that results from Rindler coordinates only covering a wedge of Minkowski spacetime.
21 As 21 As ^(21)As{ }^{21} \mathrm{As}21As in our previous discussions of Rindler spacetime, we describe Minkowski spacetime with coordinates ( T , X ) ( T , X ) (T,X)(T, X)(T,X) and line element d s 2 = d T 2 + d s 2 = d T 2 + ds^(2)=-dT^(2)+\mathrm{d} s^{2}=-\mathrm{d} T^{2}+ds2=dT2+ d X 2 d X 2 dX^(2)\mathrm{d} X^{2}dX2.

Exercises

(28.1) Compute the area of the event horizon of a (28.4) (a) Show that near the event horizon, the Schwarzschild black hole.
(28.2) Show that the heat capacity C C CCC of a Schwarzschild black hole is given by
(28.23) C = T d S d T = 2 16 π k B T 2 (28.23) C = T d S d T = 2 16 π k B T 2 {:(28.23)C=T((d)S)/((d)T)=-(ℏ^(2))/(16 pik_(B)T^(2)):}\begin{equation*} C=T \frac{\mathrm{~d} S}{\mathrm{~d} T}=-\frac{\hbar^{2}}{16 \pi k_{\mathrm{B}} T^{2}} \tag{28.23} \end{equation*}(28.23)C=T dS dT=216πkBT2
Explain why the minus sign makes physical sense.
(28.3) We shall carry out the computation sketched in Example 28.4
(a) Using incoming Eddington-Finkelstein coordinates, show that the vector with components ξ μ = ( 1 , 0 , 0 , 0 ) ξ μ = ( 1 , 0 , 0 , 0 ) xi^(mu)=(1,0,0,0)\xi^{\mu}=(1,0,0,0)ξμ=(1,0,0,0) corresponds to the 1 -form with components
(28.24) ξ μ = ( 1 + 2 M r , 1 , 0 , 0 ) (28.24) ξ μ = 1 + 2 M r , 1 , 0 , 0 {:(28.24)xi_(mu)=(-1+(2M)/(r),1,0,0):}\begin{equation*} \xi_{\mu}=\left(-1+\frac{2 M}{r}, 1,0,0\right) \tag{28.24} \end{equation*}(28.24)ξμ=(1+2Mr,1,0,0)
(b) All Killing vectors obey Killing's equation ξ μ ; ν + ξ ν ; μ = 0 ξ μ ; ν + ξ ν ; μ = 0 xi_(mu;nu)+xi_(nu;mu)=0\xi_{\mu ; \nu}+\xi_{\nu ; \mu}=0ξμ;ν+ξν;μ=0. Use this to show that an alternative expression for the surface gravity is
(28.25) ξ μ ξ μ ; σ = κ ξ σ . (28.25) ξ μ ξ μ ; σ = κ ξ σ . {:(28.25)-xi^(mu)xi_(mu)^(;sigma)=kappaxi^(sigma).:}\begin{equation*} -\xi^{\mu} \xi_{\mu}^{; \sigma}=\kappa \xi^{\sigma} . \tag{28.25} \end{equation*}(28.25)ξμξμ;σ=κξσ.
(c) Hence, show that the differential equation we need to solve is
(28.26) ξ V ; r = κ (28.26) ξ V ; r = κ {:(28.26)xi_(V;r)=-kappa:}\begin{equation*} \xi_{V ; r}=-\kappa \tag{28.26} \end{equation*}(28.26)ξV;r=κ
(d) The necessary connection coefficients are Γ V r V = 0 Γ V r V = 0 Gamma^(V)_(rV)=0\Gamma^{V}{ }_{r V}=0ΓVrV=0 and Γ r V r = M / r 2 Γ r V r = M / r 2 Gamma_(rV)^(r)=-M//r^(2)\Gamma_{r V}^{r}=-M / r^{2}ΓrVr=M/r2. Use these and the rule
(28.27) X α ; μ = X α , μ Γ α μ σ X σ (28.27) X α ; μ = X α , μ Γ α μ σ X σ {:(28.27)X_(alpha;mu)=X_(alpha,mu)-Gamma_(alpha mu)^(sigma)X_(sigma):}\begin{equation*} X_{\alpha ; \mu}=X_{\alpha, \mu}-\Gamma_{\alpha \mu}^{\sigma} X_{\sigma} \tag{28.27} \end{equation*}(28.27)Xα;μ=Xα,μΓαμσXσ
to obtain eqn 28.15.
Schwarzschild metric takes the form
(28.28) d s 2 = ( r 2 M ) 2 M d t 2 + 2 M r 2 M d r 2 + 4 M 2 d Ω 2 (28.28) d s 2 = ( r 2 M ) 2 M d t 2 + 2 M r 2 M d r 2 + 4 M 2 d Ω 2 {:(28.28)ds^(2)=-((r-2M))/(2M)dt^(2)+(2M)/(r-2M)dr^(2)+4M^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\frac{(r-2 M)}{2 M} \mathrm{~d} t^{2}+\frac{2 M}{r-2 M} \mathrm{~d} r^{2}+4 M^{2} \mathrm{~d} \Omega^{2} \tag{28.28} \end{equation*}(28.28)ds2=(r2M)2M dt2+2Mr2M dr2+4M2 dΩ2
with d Ω 2 = d θ + sin 2 θ d ϕ 2 d Ω 2 = d θ + sin 2 θ d ϕ 2 dOmega^(2)=dtheta+sin^(2)thetadphi^(2)\mathrm{d} \Omega^{2}=\mathrm{d} \theta+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ2=dθ+sin2θ dϕ2.
(b) By changing variables to ρ 2 = 8 M ( r 2 M ) ρ 2 = 8 M ( r 2 M ) rho^(2)=8M(r-2M)\rho^{2}=8 M(r-2 M)ρ2=8M(r2M), show that we obtain a line element
(28.29) d s 2 = ρ 2 16 M 2 d t 2 + d ρ 2 + 4 M 2 d Ω 2 (28.29) d s 2 = ρ 2 16 M 2 d t 2 + d ρ 2 + 4 M 2 d Ω 2 {:(28.29)ds^(2)=-(rho^(2))/(16M^(2))dt^(2)+drho^(2)+4M^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2}=-\frac{\rho^{2}}{16 M^{2}} \mathrm{~d} t^{2}+\mathrm{d} \rho^{2}+4 M^{2} \mathrm{~d} \Omega^{2} \tag{28.29} \end{equation*}(28.29)ds2=ρ216M2 dt2+dρ2+4M2 dΩ2
Readers with a background in quantum field theory will be aware that to compute thermodynamic effects in a field theory we must transform from evolution in time to evolution in Euclidean time, by making a Wick rotation. This takes the time evolution from coordinate time to a periodic imaginary time t E t E t_(E)t_{\mathrm{E}}tE which in turn corresponds to a temperature T T TTT via i t E = / k B T i t E = / k B T it_(E)=ℏ//k_(B)T\mathrm{i} t_{\mathrm{E}}=\hbar / k_{\mathrm{B}} TitE=/kBT. See the book by Zee for further discussion.
(c) Carry out the Wick rotation by setting t = t = t=t=t= i t E i t E -it_(E)-\mathrm{i} t_{\mathrm{E}}itE and make a change of variable t E = 4 M ψ t E = 4 M ψ t_(E)=4M psit_{\mathrm{E}}=4 M \psitE=4Mψ to obtain the line element
(28.30) d s 2 d ρ 2 + ρ 2 d ψ 2 + 4 M 2 d Ω 2 (28.30) d s 2 d ρ 2 + ρ 2 d ψ 2 + 4 M 2 d Ω 2 {:(28.30)ds^(2)~~drho^(2)+rho^(2)dpsi^(2)+4M^(2)dOmega^(2):}\begin{equation*} \mathrm{d} s^{2} \approx \mathrm{~d} \rho^{2}+\rho^{2} \mathrm{~d} \psi^{2}+4 M^{2} \mathrm{~d} \Omega^{2} \tag{28.30} \end{equation*}(28.30)ds2 dρ2+ρ2 dψ2+4M2 dΩ2
For fixed Ω Ω Omega\OmegaΩ this is simply flat space in cylindrical coordinates ( ρ , ψ ) ( ρ , ψ ) (rho,psi)(\rho, \psi)(ρ,ψ). The final term puts a 2-sphere of radius 2 M 2 M 2M2 M2M at every point in spacetime.
(d) Interpreting ψ ψ psi\psiψ as an angle, identify the period of t E t E t_(E)t_{\mathrm{E}}tE and determine the Hawking temperature near the horizon of the black hole.

Charged and rotating black holes

The finite is annihilated in the presence of the infinite, and becomes a pure nothing. So our spirit before God, so our justice before divine justice.
Blaise Pascal (1623-1662)
The Schwarzschild black hole is only the simplest of a more general class of objects featuring in solutions to the Einstein equation describing the spacetime around a distribution of mass. In this chapter, we examine the consequence of (i) the mass carrying electrical charge and (ii) the mass rotating. We shall assume that gravitational collapse of the mass distribution exposes the region of space close to the origin, so that a black hole is created. In both cases, we shall encounter singularities along with a rich seam of new physics.
Owing to the presence of an event horizon, a black hole is necessarily hidden from the view of external observers. The only properties of the black hole that we observers should expect to be able to access are the mass, charge and angular momentum. Further, owing to the singularity, it is likely that these three properties alone describe the metric field near the hole. As a result, the properties of the black-hole spacetime are remarkably independent of the history of the spacetime, such as the matter content that has fallen into the hole. This state of affairs was described by the slogan 'black holes have no hair' by John Wheeler 1 1 ^(1){ }^{1}1 and consequently elevated to a mathematical no-hair theorem, using hair as a metaphor for all other information in the spacetime. As a consequence of this, we expect our discussion to be widely applicable to black holes in the Universe, however they have arisen.

29.1 Charged black holes

In most astrophysical contexts, it is unlikely that an electrically charged black hole could be stabilized, since we would expect that it would attract matter with the opposite electrical charge and eventually become charge neutral. As we shall see, the consequence of allowing a black hole to have a charge is rather interesting. 2 2 ^(2){ }^{2}2
Suspending our disbelief, we therefore consider an electrically charged distribution of mass M M MMM with total charge q q qqq. The presence of the charge means that the solution to the Einstein equation must allow for the right-
29.1 Charged black holes 297 29.2 Kerr black holes 299 29.3 Interacting with the Kerr geometry 304
Chapter summary 305
Exercises 306
1 1 ^(1){ }^{1}1 Wheeler reported that his student Jacob Bekenstein (1947-2015) had invented the phrase. The justification of the theorem was due to Stephen Hawking, Brandon Carter and Werner Israel (1931-2022).
2 2 ^(2){ }^{2}2 The physics is described by the Reissner-Nordström line element which we will now introduce. Hans Reissher (1874-1967), an aeronautical engineer, first discovered this solution. It was also found by: (i) Gunnar Nordström (1881-1923), described as the 'Finnish Einstein', who ultimately developed an alternative theory of gravitation to general relativity; (ii) Hermann Weyl (1885-1955), the genius who made significant advances across large numbers of topics in mathematical physics including general relativity; and (iii) George Barker Jeffery (18911957), who was famed in Britain as a translator of the original papers on relativity into the English language.
3 3 ^(3){ }^{3}3 The possibility of this configuration realizing a black hole might lead one to wonder if the metric describes the spacetime close to an electron, which after all, is an infinitely dense, charged object. This turns out not to be the case since the electronic spin is not taken into account in our description.
Fig. 29.1 Penrose diagram for a charged black hole. Dotted lines are r = r + r = r + r=r_(+)r=r_{+}r=r+; dashed lines are r = r r = r r=r_(-)r=r_{-}r=r; doubled lines are r = 0 r = 0 r=0r=0r=0.
4 4 ^(4){ }^{4}4 Recall that the singularity at t = 0 t = 0 t=0t=0t=0 in the Robertson-Walker Penrose diagram is spacelike, as is the r = 0 r = 0 r=0r=0r=0 singularity in the Schwarzschild black hole, owing to the swap between the roles of space and time inside the event horizon This is the first timelike singularity that we have encountered. It is depicted by we have encountered. It is depicted by
a double vertical line in the Penrose dia double
agram.
hand side to feature electromagnetic energy density. This results in an upgraded metric field that, in the limit q 0 q 0 q rarr0q \rightarrow 0q0, gives the Schwarzschild line element. 3 3 ^(3){ }^{3}3
The Reissner-Nordström line element describes the metric field outside of the electrically charged, spherically symmetric mass distribution. It is given by
d s 2 = ( 1 2 M r + q 2 r 2 ) d t 2 + ( 1 2 M r + q 2 r 2 ) 1 d r 2 (29.1) + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = 1 2 M r + q 2 r 2 d t 2 + 1 2 M r + q 2 r 2 1 d r 2 (29.1) + r 2 d θ 2 + sin 2 θ d ϕ 2 {:[ds^(2)=-(1-(2M)/(r)+(q^(2))/(r^(2)))dt^(2)+(1-(2M)/(r)+(q^(2))/(r^(2)))^(-1)dr^(2)],[(29.1)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))]:}\begin{align*} \mathrm{d} s^{2}= & -\left(1-\frac{2 M}{r}+\frac{q^{2}}{r^{2}}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}+\frac{q^{2}}{r^{2}}\right)^{-1} \mathrm{~d} r^{2} \\ & +r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{29.1} \end{align*}ds2=(12Mr+q2r2)dt2+(12Mr+q2r2)1 dr2(29.1)+r2( dθ2+sin2θ dϕ2)
If q 2 > M 2 q 2 > M 2 q^(2) > M^(2)q^{2}>M^{2}q2>M2 the metric is non-singular everywhere, except for r = 0 r = 0 r=0r=0r=0. If q 2 M 2 q 2 M 2 q^(2) <= M^(2)q^{2} \leq M^{2}q2M2 there are coordinate singularities at radii r + r + r_(+)r_{+}r+and r r r_(-)r_{-}r, where r ± = M ± ( M 2 q 2 ) 1 2 r ± = M ± M 2 q 2 1 2 r_(+-)=M+-(M^(2)-q^(2))^((1)/(2))r_{ \pm}=M \pm\left(M^{2}-q^{2}\right)^{\frac{1}{2}}r±=M±(M2q2)12
Using these singularities, we identify three regions of the spacetime. Region I takes the large values of the radial coordinate: r + < r < r + < r < r_(+) < r < oor_{+}<r<\inftyr+<r<; region II has r < r < r + r < r < r + r_(-) < r < r_(+)r_{-}<r<r_{+}r<r<r+; and region III is the inner region with 0 < r < r 0 < r < r 0 < r < r_(-)0<r<r_{-}0<r<r. The first and third regions are static; region II is not. (Note that if q 2 = M 2 q 2 = M 2 q^(2)=M^(2)q^{2}=M^{2}q2=M2 then region II does not exist.)
Maximally extended Penrose diagrams for the charged black hole are shown in Figs. 29.1 and 29.2, where regions I, II and III are indicated. Region I is asymptotically flat and therefore features null, spacelike and timelike infinities. There are an infinite number of these regions I in the Penrose diagram, all connected by regions II and III (we shall see why shortly). In region I, surfaces of constant t t ttt are spacelike and surfaces of constant r r rrr are timelike, in the usual manner. In region II, surfaces of constant r r rrr are spacelike and surfaces of constant t t ttt timelike, just as we have in the region II of a Schwarzschild black hole. Each point in region II therefore represents a closed trapped surface.
In region III, all lines of constant r r rrr are timelike and lines of constant t t ttt are spacelike, just as in region I (Fig. 29.2). Region III has a physical singularity at r = 0 r = 0 r=0r=0r=0. but, owing to the different roles of space and time compared to the Schwarzschild case, it is not spacelike but timelike. 4 4 ^(4){ }^{4}4 The singularity in region III is not only timelike but is also repulsive, since no timelike geodesic intersects it and so freely falling massive particles will not collide with it. The singularity can then be avoided by a timelike curve that crosses the surface r + r + r_(+)r_{+}r+from region I into region II and then into region III. A timelike curve can then pass into another region II and into a further (asymptotically flat) region I. The traveller following this curve will have tunnelled into a new Universe through the wormhole caused by the presence of the charge! This is a one-way process, so the traveller will not return, since there is no future-directed timelike curve that allows the traveller back to the original region-I universe.
What would observers see in this spacetime? As in the Schwarzschild case, an observer close to infinity would see an astronaut infinitely redshifted as she approaches the surface r + r + r_(+)r_{+}r+. From the geometry in Fig. 29.1, an observer crossing surface r r r_(-)r_{-}rinto region III would potentially be able
to see the whole history of a region I in a finite time. Light from objects in region I would appear infinitely blueshifted as the astronaut approached i + i + i^(+)i^{+}i+.

29.2 Kerr black holes

We have done a lot using the Schwarzschild solution, but now there's some bad news: most astronomical objects rotate and so are not described by the Schwarzschild geometry, which assumes spherical symmetry (which is a symmetry that is broken by rotation). The Kerr geometry 5 5 ^(5){ }^{5}5 is the only known exact solution to the Einstein equation that is able to represent a stationary, axially symmetric, asymptotically flat spacetime outside of a rotating distribution of mass.

Example 29.1

So far, most of our geometries have been expressible in terms of a diagonal metric. 6 6 ^(6){ }^{6}6 The (3+1)-dimensional metric describing the rotating Kerr geometry is stationary (i.e. none of the metric components depend on t t ttt ) and axisymmetric, rather than spherically symmetric, such that we expect that there is no dependence of the components on ϕ ϕ phi\phiϕ. The most general form of such a stationary, axisymmetric line element is often written in terms of an angular velocity ω ω omega\omegaω as
(29.2) d s 2 = e 2 ν d t 2 + e 2 ψ ( d ϕ ω d t ) 2 + e 2 μ 2 ( d x 2 ) 2 + e 2 μ 3 ( d x 3 ) 2 (29.2) d s 2 = e 2 ν d t 2 + e 2 ψ ( d ϕ ω d t ) 2 + e 2 μ 2 d x 2 2 + e 2 μ 3 d x 3 2 {:(29.2)ds^(2)=-e^(2nu)dt^(2)+e^(2psi)(dphi-omegadt)^(2)+e^(2mu_(2))((d)x^(2))^(2)+e^(2mu_(3))((d)x^(3))^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{e}^{2 \nu} \mathrm{~d} t^{2}+\mathrm{e}^{2 \psi}(\mathrm{~d} \phi-\omega \mathrm{d} t)^{2}+\mathrm{e}^{2 \mu_{2}}\left(\mathrm{~d} x^{2}\right)^{2}+\mathrm{e}^{2 \mu_{3}}\left(\mathrm{~d} x^{3}\right)^{2} \tag{29.2} \end{equation*}(29.2)ds2=e2ν dt2+e2ψ( dϕωdt)2+e2μ2( dx2)2+e2μ3( dx3)2
where ν , ψ , ω , μ 2 ν , ψ , ω , μ 2 nu,psi,omega,mu_(2)\nu, \psi, \omega, \mu_{2}ν,ψ,ω,μ2, and μ 3 μ 3 mu_(3)\mu_{3}μ3 are functions of variables x 2 x 2 x^(2)x^{2}x2 and x 3 x 3 x^(3)x^{3}x3. Notice that the second term implies that the resulting metric will not be diagonal, but will include a cross term proportional to ω d ϕ d t ω d ϕ d t omegadphidt\omega \mathrm{d} \phi \mathrm{d} tωdϕdt. We shall see later that the cross term leads to lots of the notable properties of this geometry.
The Kerr geometry is usually written in yet another new coordinate system: the so-called Boyer-Lindquist coordinates 7 ( t , r , θ , ϕ ) 7 ( t , r , θ , ϕ ) ^(7)(t,r,theta,phi){ }^{7}(t, r, \theta, \phi)7(t,r,θ,ϕ). Although these seem to resemble ordinary spherical polars, they are slightly unusual. In particular, although t t ttt and ϕ ϕ phi\phiϕ are the same as their counterparts in standard spherical polars; θ θ theta\thetaθ and r r rrr are not. These latter two coordinates are related to Cartesian coordinates via
x = ( r 2 + a 2 ) 1 2 sin θ cos ϕ (29.3) y = ( r 2 + a 2 ) 1 2 sin θ sin ϕ x = r 2 + a 2 1 2 sin θ cos ϕ (29.3) y = r 2 + a 2 1 2 sin θ sin ϕ {:[x=(r^(2)+a^(2))^((1)/(2))sin theta cos phi],[(29.3)y=(r^(2)+a^(2))^((1)/(2))sin theta sin phi]:}\begin{align*} & x=\left(r^{2}+a^{2}\right)^{\frac{1}{2}} \sin \theta \cos \phi \\ & y=\left(r^{2}+a^{2}\right)^{\frac{1}{2}} \sin \theta \sin \phi \tag{29.3} \end{align*}x=(r2+a2)12sinθcosϕ(29.3)y=(r2+a2)12sinθsinϕ
Notice that these coordinates are a little like spherical polars with the radius adjusted. In terms of these coordinates, a surface of constant r r rrr is no longer a sphere, but rather is an ellipsoid. (Setting r = 0 r = 0 r=0r=0r=0 does not, therefore, result in a point, so the physical singularity for a Kerr black hole is more complex than in the cases examined so far.)
In Boyer-Lindquist coordinates, we can express the metric field for the Kerr geometry as follows:
Fig. 29.2 Penrose diagram for the charged hole, showing surfaces of constant r r rrr (solid lines) and surfaces of constant t t ttt (dashed lines).
5 5 ^(5){ }^{5}5 Roy Kerr (1934- ) is, in addition to his mathematical achievements, noted as a bridge player, having represented New Zealand in the 1970s. The generalization of the metric for a charged, rotating mass is called the Kerr-Newman geometry, named also after Ezra 'Ted' Newman (1929-2021).
6 6 ^(6){ }^{6}6 In fact, the Cotton-Darboux theorem guarantees that the metric g = g = g=g=g= g μ ν e μ e ν g μ ν e μ e ν g^(mu nu)e_(mu)oxe_(nu)g^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu}gμνeμeν, with μ , ν = 0 , 1 , 2 μ , ν = 0 , 1 , 2 mu,nu=0,1,2\mu, \nu=0,1,2μ,ν=0,1,2 in a three-dimensional space, can always be brought into diagonal form by a local coordinate transformation. See Chandrasekhar for the proof.
7 7 ^(7){ }^{7}7 Robert H. Boyer (1932-1966) and Richard W. Lindquist. Their paper on these coordinates was published in 1967, posthumously for Boyer, who was tragically killed in the infamous University of Texas tower shooting.
The Kerr metric has components
g t t = ( 1 2 M r ρ 2 ) = Δ a 2 sin 2 θ ρ 2 , g t ϕ = 2 M a r sin 2 θ ρ 2 , g ϕ ϕ = ( r 2 + a 2 ) 2 a 2 Δ sin 2 θ ρ 2 sin 2 θ , g r r = ρ 2 Δ , (29.5) g θ θ = ρ 2 . g t t = 1 2 M r ρ 2 = Δ a 2 sin 2 θ ρ 2 , g t ϕ = 2 M a r sin 2 θ ρ 2 , g ϕ ϕ = r 2 + a 2 2 a 2 Δ sin 2 θ ρ 2 sin 2 θ , g r r = ρ 2 Δ , (29.5) g θ θ = ρ 2 . {:[g_(tt)=-(1-(2Mr)/(rho^(2)))],[=-(Delta-a^(2)sin^(2)theta)/(rho^(2))","],[g_(t phi)=-(2Marsin^(2)theta)/(rho^(2))","],[g_(phi phi)=((r^(2)+a^(2))^(2)-a^(2)Deltasin^(2)theta)/(rho^(2))*sin^(2)theta","],[g_(rr)=(rho^(2))/(Delta)","],[(29.5)g_(theta theta)=rho^(2).]:}\begin{align*} g_{t t} & =-\left(1-\frac{2 M r}{\rho^{2}}\right) \\ & =-\frac{\Delta-a^{2} \sin ^{2} \theta}{\rho^{2}}, \\ g_{t \phi} & =-\frac{2 M a r \sin ^{2} \theta}{\rho^{2}}, \\ g_{\phi \phi} & =\frac{\left(r^{2}+a^{2}\right)^{2}-a^{2} \Delta \sin ^{2} \theta}{\rho^{2}} \cdot \sin ^{2} \theta, \\ g_{r r} & =\frac{\rho^{2}}{\Delta}, \\ g_{\theta \theta} & =\rho^{2} . \tag{29.5} \end{align*}gtt=(12Mrρ2)=Δa2sin2θρ2,gtϕ=2Marsin2θρ2,gϕϕ=(r2+a2)2a2Δsin2θρ2sin2θ,grr=ρ2Δ,(29.5)gθθ=ρ2.
We also have Killing vectors ξ ξ xi\boldsymbol{\xi}ξ and η η eta\boldsymbol{\eta}η with components
ξ μ = ( 1 , 0 , 0 , 0 ) η μ = ( 0 , 0 , 0 , 1 ) ξ μ = ( 1 , 0 , 0 , 0 ) η μ = ( 0 , 0 , 0 , 1 ) {:[xi^(mu)=(1","0","0","0)],[eta^(mu)=(0","0","0","1)]:}\begin{aligned} \xi^{\mu} & =(1,0,0,0) \\ \eta^{\mu} & =(0,0,0,1) \end{aligned}ξμ=(1,0,0,0)ημ=(0,0,0,1)
We therefore have conserved quantities
E ~ = ξ u = u t L ~ = η u = u ϕ . E ~ = ξ u = u t L ~ = η u = u ϕ . {:[ tilde(E)=-xi*u=-u_(t)],[ tilde(L)=eta*u=u_(phi).]:}\begin{gathered} \tilde{E}=-\boldsymbol{\xi} \cdot \boldsymbol{u}=-u_{t} \\ \tilde{L}=\boldsymbol{\eta} \cdot \boldsymbol{u}=u_{\phi} . \end{gathered}E~=ξu=utL~=ηu=uϕ.
and
Note that, in the spirit of eqn 29.2, the line element can be rewritten as d s 2 = ρ 2 Δ Σ 2 d t 2 + Σ 2 sin 2 θ ρ 2 ( d ϕ ω d t ) 2 d s 2 = ρ 2 Δ Σ 2 d t 2 + Σ 2 sin 2 θ ρ 2 ( d ϕ ω d t ) 2 ds^(2)=-(rho^(2)Delta)/(Sigma^(2))*dt^(2)+(Sigma^(2)sin^(2)theta)/(rho^(2))(dphi-omegadt)^(2)\mathrm{d} s^{2}=-\frac{\rho^{2} \Delta}{\Sigma^{2}} \cdot \mathrm{~d} t^{2}+\frac{\Sigma^{2} \sin ^{2} \theta}{\rho^{2}}(\mathrm{~d} \phi-\omega \mathrm{d} t)^{2}ds2=ρ2ΔΣ2 dt2+Σ2sin2θρ2( dϕωdt)2 + ρ 2 Δ 2 d r 2 + ρ 2 d θ 2 + ρ 2 Δ 2 d r 2 + ρ 2 d θ 2 +(rho^(2))/(Delta^(2))dr^(2)+rho^(2)dtheta^(2)+\frac{\rho^{2}}{\Delta^{2}} \mathrm{~d} r^{2}+\rho^{2} \mathrm{~d} \theta^{2}+ρ2Δ2 dr2+ρ2 dθ2,
where Σ 2 = ( r 2 + a 2 ) 2 a 2 Δ sin 2 θ Σ 2 = r 2 + a 2 2 a 2 Δ sin 2 θ Sigma^(2)=(r^(2)+a^(2))^(2)-a^(2)Deltasin^(2)theta\Sigma^{2}=\left(r^{2}+a^{2}\right)^{2}-a^{2} \Delta \sin ^{2} \thetaΣ2=(r2+a2)2a2Δsin2θ and ω = 2 Mar / Σ 2 ω = 2 Mar / Σ 2 omega=2Mar//Sigma^(2)\omega=2 \mathrm{Mar} / \Sigma^{2}ω=2Mar/Σ2.
d s 2 = ( 1 2 M r ρ 2 ) d t 2 4 M a r sin 2 θ ρ 2 d t d ϕ + ( r 2 + a 2 ) 2 a 2 Δ sin 2 θ ρ 2 sin 2 θ d ϕ 2 (29.4) + ρ 2 Δ d r 2 + ρ 2 d θ 2 , d s 2 = 1 2 M r ρ 2 d t 2 4 M a r sin 2 θ ρ 2 d t d ϕ + r 2 + a 2 2 a 2 Δ sin 2 θ ρ 2 sin 2 θ d ϕ 2 (29.4) + ρ 2 Δ d r 2 + ρ 2 d θ 2 , {:[ds^(2)=-(1-(2Mr)/(rho^(2)))dt^(2)-(4Marsin^(2)theta)/(rho^(2))dtdphi],[+((r^(2)+a^(2))^(2)-a^(2)Deltasin^(2)theta)/(rho^(2))sin^(2)thetadphi^(2)],[(29.4)+(rho^(2))/(Delta)dr^(2)+rho^(2)dtheta^(2)","]:}\begin{align*} \mathrm{d} s^{2}= & -\left(1-\frac{2 M r}{\rho^{2}}\right) \mathrm{d} t^{2}-\frac{4 M a r \sin ^{2} \theta}{\rho^{2}} \mathrm{~d} t \mathrm{~d} \phi \\ & +\frac{\left(r^{2}+a^{2}\right)^{2}-a^{2} \Delta \sin ^{2} \theta}{\rho^{2}} \sin ^{2} \theta \mathrm{~d} \phi^{2} \\ & +\frac{\rho^{2}}{\Delta} \mathrm{~d} r^{2}+\rho^{2} \mathrm{~d} \theta^{2}, \tag{29.4} \end{align*}ds2=(12Mrρ2)dt24Marsin2θρ2 dt dϕ+(r2+a2)2a2Δsin2θρ2sin2θ dϕ2(29.4)+ρ2Δ dr2+ρ2 dθ2,
where Δ = r 2 2 M r + a 2 Δ = r 2 2 M r + a 2 Delta=r^(2)-2Mr+a^(2)\Delta=r^{2}-2 M r+a^{2}Δ=r22Mr+a2, and ρ 2 = r 2 + a 2 cos 2 θ ρ 2 = r 2 + a 2 cos 2 θ rho^(2)=r^(2)+a^(2)cos^(2)theta\rho^{2}=r^{2}+a^{2} \cos ^{2} \thetaρ2=r2+a2cos2θ.
Here M M MMM is the mass of the distribution and J = M a J = M a J=MaJ=M aJ=Ma is its angular momentum as measured by an observer at infinity. The parameter a a aaa, which has units of length, is known as the Kerr parameter. It is a measure of the rotation.
The Kerr geometry is invariant under the simultaneous transformation t t t t t rarr-tt \rightarrow-ttt and ϕ ϕ ϕ ϕ phi rarr-phi\phi \rightarrow-\phiϕϕ, just as we should expect for any axially rotating object (i.e. if we reverse the direction of time, the mass rotates in the opposite direction). In the limit of large r r rrr we recover Minkowski space, confirming that the line element is asymptotically flat. When a 0 a 0 a rarr0a \rightarrow 0a0, we recover the Schwarzschild geometry. By inspection of the Killing vectors, we have, just as for the Schwarzschild geometry, conservation of u t u t u_(t)u_{t}ut and u ϕ u ϕ u_(phi)u_{\phi}uϕ (although the presence of cross terms in the metric complicates the computation of u t u t u^(t)u^{t}ut and u ϕ u ϕ u^(phi)u^{\phi}uϕ ).
The anatomy of this spacetime is very rich in features and is shown in cartoon form in Fig. 29.3. There are two sets of singularities in this geometry: (i) ρ = 0 ρ = 0 rho=0\rho=0ρ=0 and (ii) Δ = 0 Δ = 0 Delta=0\Delta=0Δ=0. The former is a class of physical singularity, while the latter are coordinate singularities. Substituting into eqn 29.3, we find that ρ = 0 ρ = 0 rho=0\rho=0ρ=0 corresponds to a ring of Cartesian coordinates with x 2 + y 2 = a 2 x 2 + y 2 = a 2 x^(2)+y^(2)=a^(2)x^{2}+y^{2}=a^{2}x2+y2=a2 and z = 0 z = 0 z=0z=0z=0 (or, equivalently, θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 ). These are physical singularities at the heart of the Kerr black hole where the curvature of spacetime is infinite.
When the parameters satisfy 0 a 2 M 2 0 a 2 M 2 0 <= a^(2) <= M^(2)0 \leq a^{2} \leq M^{2}0a2M2 (the case we will examine exclusively in this section) we have a situation where three regions of interest are separated by singular surfaces r ± r ± r_(+-)r_{ \pm}r±(Fig. 29.3), not unlike the Reissner-Nordström black hole. Here the coordinate singularities occur where Δ Δ Delta\DeltaΔ vanishes at values of r r rrr given by
(29.7) r ± = M ± ( M 2 a 2 ) 1 2 (29.7) r ± = M ± M 2 a 2 1 2 {:(29.7)r_(+-)=M+-(M^(2)-a^(2))^((1)/(2)):}\begin{equation*} r_{ \pm}=M \pm\left(M^{2}-a^{2}\right)^{\frac{1}{2}} \tag{29.7} \end{equation*}(29.7)r±=M±(M2a2)12
These equations describe spherical surfaces, both of which behave as event horizons. In Fig. 29.3, they are labelled as the inner ( r ) r (r_(-))\left(r_{-}\right)(r)and outer ( r + ) r + (r_(+))\left(r_{+}\right)(r+)event horizons. Note that when a = 0 a = 0 a=0a=0a=0 we have that the outer event horizon has r + = 2 M r + = 2 M r_(+)=2Mr_{+}=2 Mr+=2M, while the inner event horizon r r r_(-)r_{-}rvanishes, and so we recover the Schwarzschild case. The presence of the square root in eqn 29.7 shows that the event horizons only exists for a M a M a <= Ma \leq MaM, and so the hole has a maximum angular momentum J = M 2 J = M 2 J=M^(2)J=M^{2}J=M2. Black holes with this maximum angular momentum are known as extremal Kerr black holes.
Fig. 29.3 The singular surfaces in the Kerr geometry.

Example 29.2

To compute the area of the r + r + r_(+)r_{+}r+horizon in the Kerr black hole, we set d t = d r = 0 d t = d r = 0 dt=dr=0\mathrm{d} t=\mathrm{d} r=0dt=dr=0 and substitute r = r + r = r + r=r_(+)r=r_{+}r=r+and ρ ( r + ) ρ r + rho(r_(+))\rho\left(r_{+}\right)ρ(r+)to obtain
d s 2 = ρ + 2 d θ 2 + ( r + 2 + a 2 + 2 M r + a 2 sin 2 θ ρ + 2 ) sin 2 θ d ϕ 2 (29.8) = ρ + 2 d θ 2 + ( 2 M r + ρ + ) 2 sin 2 θ d ϕ 2 d s 2 = ρ + 2 d θ 2 + r + 2 + a 2 + 2 M r + a 2 sin 2 θ ρ + 2 sin 2 θ d ϕ 2 (29.8) = ρ + 2 d θ 2 + 2 M r + ρ + 2 sin 2 θ d ϕ 2 {:[ds^(2)=rho_(+)^(2)dtheta^(2)+(r_(+)^(2)+a^(2)+(2Mr_(+)a^(2)sin^(2)theta)/(rho_(+)^(2)))sin^(2)thetadphi^(2)],[(29.8)=rho_(+)^(2)dtheta^(2)+((2Mr_(+))/(rho_(+)))^(2)sin^(2)thetadphi^(2)]:}\begin{align*} \mathrm{d} s^{2} & =\rho_{+}^{2} \mathrm{~d} \theta^{2}+\left(r_{+}^{2}+a^{2}+\frac{2 M r_{+} a^{2} \sin ^{2} \theta}{\rho_{+}^{2}}\right) \sin ^{2} \theta \mathrm{~d} \phi^{2} \\ & =\rho_{+}^{2} \mathrm{~d} \theta^{2}+\left(\frac{2 M r_{+}}{\rho_{+}}\right)^{2} \sin ^{2} \theta \mathrm{~d} \phi^{2} \tag{29.8} \end{align*}ds2=ρ+2 dθ2+(r+2+a2+2Mr+a2sin2θρ+2)sin2θ dϕ2(29.8)=ρ+2 dθ2+(2Mr+ρ+)2sin2θ dϕ2
where ρ + 2 = r + 2 + a 2 cos 2 θ ρ + 2 = r + 2 + a 2 cos 2 θ rho_(+)^(2)=r_(+)^(2)+a^(2)cos^(2)theta\rho_{+}^{2}=r_{+}^{2}+a^{2} \cos ^{2} \thetaρ+2=r+2+a2cos2θ and r + r + r_(+)r_{+}r+is defined by r + 2 + a 2 = 2 M r + r + 2 + a 2 = 2 M r + r_(+)^(2)+a^(2)=2Mr_(+)r_{+}^{2}+a^{2}=2 M r_{+}r+2+a2=2Mr+. The area of the Kerr horizon is then
(29.9) A = 0 2 π d ϕ 0 π d θ g = 8 π M r + , (29.9) A = 0 2 π d ϕ 0 π d θ g = 8 π M r + , {:(29.9)A=int_(0)^(2pi)dphiint_(0)^(pi)dtheta g=8pi Mr_(+)",":}\begin{equation*} A=\int_{0}^{2 \pi} \mathrm{~d} \phi \int_{0}^{\pi} \mathrm{d} \theta g=8 \pi M r_{+}, \tag{29.9} \end{equation*}(29.9)A=02π dϕ0πdθg=8πMr+,
where g g ggg is the determinant of the matrix of metric components. The integral is examined further in the exercises.
The Penrose diagram for the Kerr black hole is shown in Fig. 29.4. In this case, the maximal extension of the spacetime allows us to avoid the ring of physical singularities, since they do not occur at r = 0 r = 0 r=0r=0r=0. As a result, we can extend spacetime into a region of negative r r rrr, in order to show the surfaces on which null lines (i.e. light rays) terminate. As in the Reissner-Nordström case, we can characterize the three regions: an asymptotically flat region I, region II containing closed trapped surfaces and region III containing the singularities. The Penrose diagram shows that a traveller can fall from region I, through the event horizon at r + r + r_(+)r_{+}r+, to find themselves in region II, where the closed, trapped surfaces occur. The traveller is forced into region III which contains the singularities.
Fig. 29.4 Penrose diagram for the Kerr geometry, showing lines of constant r r rrr (dashed). The solid curved lines (in region III) show r = 0 r = 0 r=0r=0r=0, the dotted lines shows r = r + r = r + r=r_(+)r=r_{+}r=r+and the dot-dashed line is r = r r = r r=r_(-)r=r_{-}r=r
The singularities are timelike, and so can be avoided by timelike cures, just as we saw earlier. Also notable in region III is the existence of closed, timelike curves which violate causality. As in the Reissner-Nordström case, it is possible to tunnel through a wormhole to a new region II and then to a new region I.
To understand these three regions of the Kerr geometry in more detail, we shall drop some test particles into the spacetime and make use of the conserved quantities in this geometry. As mentioned above, the conserved quantities are
E ~ = u t = g t t u t + g t ϕ u ϕ , , L ~ = u ϕ = g ϕ t u t + g ϕ ϕ u ϕ . E ~ = u t = g t t u t + g t ϕ u ϕ , , L ~ = u ϕ = g ϕ t u t + g ϕ ϕ u ϕ . {:- tilde(E)=u_(t):}=g_(tt)u^(t)+g_(t phi)u^(phi),子, tilde(L)=u_(phi)=g_(phi t)u^(t)+g_(phi phi)u^(phi).\begin{align*} -\tilde{E} & =u_{t} \end{align*}=g_{t t} u^{t}+g_{t \phi} u^{\phi}, ~ 子, ~ \tilde{L}=u_{\phi}=g_{\phi t} u^{t}+g_{\phi \phi} u^{\phi} .E~=ut=gttut+gtϕuϕ, , L~=uϕ=gϕtut+gϕϕuϕ.
Example 29.3
Consider a particle dropped into the geometry with no angular momentum, so that u ϕ = 0 u ϕ = 0 u_(phi)=0u_{\phi}=0uϕ=0 initially. We will compute d ϕ / d t d ϕ / d t dphi//dt\mathrm{d} \phi / \mathrm{d} tdϕ/dt. We have
(29.11) d ϕ d t = d ϕ d τ d τ d t = u ϕ u t . (29.11) d ϕ d t = d ϕ d τ d τ d t = u ϕ u t . {:(29.11)(dphi)/((d)t)=(dphi)/((d)tau)((d)tau)/((d)t)=(u^(phi))/(u^(t)).:}\begin{equation*} \frac{\mathrm{d} \phi}{\mathrm{~d} t}=\frac{\mathrm{d} \phi}{\mathrm{~d} \tau} \frac{\mathrm{~d} \tau}{\mathrm{~d} t}=\frac{u^{\phi}}{u^{t}} . \tag{29.11} \end{equation*}(29.11)dϕ dt=dϕ dτ dτ dt=uϕut.
Recall that the cross terms in the metric give us 8 8 ^(8){ }^{8}8
u ϕ = g ϕ ϕ u ϕ + g ϕ t u t , (29.14) u t = g t t u t + g t ϕ u ϕ , (29.15) d ϕ d t = g ϕ t g t t 0 . u ϕ = g ϕ ϕ u ϕ + g ϕ t u t , (29.14) u t = g t t u t + g t ϕ u ϕ , (29.15) d ϕ d t = g ϕ t g t t 0 . {:[u^(phi)=g^(phi phi)u_(phi)+g^(phi t)u_(t)","],[(29.14)u^(t)=g^(tt)u_(t)+g^(t phi)u_(phi)","],[(29.15)(dphi)/((d)t)=(g^(phi t))/(g^(tt))!=0.]:}\begin{gather*} u^{\phi}=g^{\phi \phi} u_{\phi}+g^{\phi t} u_{t}, \\ u^{t}=g^{t t} u_{t}+g^{t \phi} u_{\phi}, \tag{29.14}\\ \frac{\mathrm{d} \phi}{\mathrm{~d} t}=\frac{g^{\phi t}}{g^{t t}} \neq 0 . \tag{29.15} \end{gather*}uϕ=gϕϕuϕ+gϕtut,(29.14)ut=gttut+gtϕuϕ,(29.15)dϕ dt=gϕtgtt0.
The particle therefore picks up some non-zero coordinate velocity in the ϕ ϕ phi\phiϕ-direction.
From the previous example, we have the remarkable result that a particle dropped in to the Kerr geometry with no angular momentum gains some by virtue of the gravitational effect of the mass. This effect is often called inertial frame dragging, reflecting the fact that the system is dragged by the geometry in the direction of the mass's rotation. We see that the effect occurs because of the presence of a cross term g ϕ t g ϕ t g^(phi t)g^{\phi t}gϕt in the metric.
While we are considering massive particles, we can compute the effective energy equation governing the possible orbits.

Example 29.4

Consider the paths of massive particles in the Kerr metric confined to the θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 plane. We obtain coordinate velocity components
u t = 1 Δ [ ( r 2 + a 2 + 2 M a 2 r ) E ~ 2 M a r L ~ ] (29.16) u ϕ = 1 Δ [ ( 1 2 M r ) L ~ + 2 M a r E ~ ] u t = 1 Δ r 2 + a 2 + 2 M a 2 r E ~ 2 M a r L ~ (29.16) u ϕ = 1 Δ 1 2 M r L ~ + 2 M a r E ~ {:[u^(t)=(1)/(Delta)[(r^(2)+a^(2)+(2Ma^(2))/(r))( tilde(E))-(2Ma)/(r)( tilde(L))]],[(29.16)u^(phi)=(1)/(Delta)[(1-(2M)/(r))( tilde(L))+(2Ma)/(r)( tilde(E))]]:}\begin{align*} u^{t} & =\frac{1}{\Delta}\left[\left(r^{2}+a^{2}+\frac{2 M a^{2}}{r}\right) \tilde{E}-\frac{2 M a}{r} \tilde{L}\right] \\ u^{\phi} & =\frac{1}{\Delta}\left[\left(1-\frac{2 M}{r}\right) \tilde{L}+\frac{2 M a}{r} \tilde{E}\right] \tag{29.16} \end{align*}ut=1Δ[(r2+a2+2Ma2r)E~2MarL~](29.16)uϕ=1Δ[(12Mr)L~+2MarE~]
Since we also have u θ = 0 u θ = 0 u^(theta)=0u^{\theta}=0uθ=0 and u 2 = 1 u 2 = 1 u^(2)=-1\boldsymbol{u}^{2}=-1u2=1 we can solve for u r u r u^(r)u^{r}ur to yield up a result similar in form to the one we obtained in the Schwarzschild geometry. This is
(29.17) E ~ 2 1 2 = 1 2 ( d r d τ ) 2 + V eff ( r , E ~ , L ~ ) , (29.17) E ~ 2 1 2 = 1 2 d r d τ 2 + V eff ( r , E ~ , L ~ ) , {:(29.17)( tilde(E)^(2)-1)/(2)=(1)/(2)(((d)r)/((d)tau))^(2)+V_(eff)(r"," tilde(E)"," tilde(L))",":}\begin{equation*} \frac{\tilde{E}^{2}-1}{2}=\frac{1}{2}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}+V_{\mathrm{eff}}(r, \tilde{E}, \tilde{L}), \tag{29.17} \end{equation*}(29.17)E~212=12( dr dτ)2+Veff(r,E~,L~),
where in the Kerr geometry, we have an effective potential 9 9 ^(9){ }^{9}9
(29.18) V eff ( r , E ~ , L ~ ) = M r + L ~ 2 a 2 ( E ~ 2 1 ) 2 r 2 M ( L ~ a E ~ ) 2 r 3 (29.18) V eff ( r , E ~ , L ~ ) = M r + L ~ 2 a 2 E ~ 2 1 2 r 2 M ( L ~ a E ~ ) 2 r 3 {:(29.18)V_(eff)(r"," tilde(E)"," tilde(L))=-(M)/(r)+( tilde(L)^(2)-a^(2)( tilde(E)^(2)-1))/(2r^(2))-(M(( tilde(L))-a( tilde(E)))^(2))/(r^(3)):}\begin{equation*} V_{\mathrm{eff}}(r, \tilde{E}, \tilde{L})=-\frac{M}{r}+\frac{\tilde{L}^{2}-a^{2}\left(\tilde{E}^{2}-1\right)}{2 r^{2}}-\frac{M(\tilde{L}-a \tilde{E})^{2}}{r^{3}} \tag{29.18} \end{equation*}(29.18)Veff(r,E~,L~)=Mr+L~2a2(E~21)2r2M(L~aE~)2r3
Next, we consider dropping a photon into the equatorial plane of the geometry, defined by θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2.
Example 29.5
Consider a photon initially travelling in the ± ϕ ± ϕ +-phi\pm \phi±ϕ direction, so that only d t d t dt\mathrm{d} tdt and d ϕ d ϕ dphi\mathrm{d} \phidϕ are non-zero. For photons, we can compute the motion using the null condition
(29.19) d s 2 = g t t d t 2 + 2 g t ϕ d t d ϕ + g ϕ ϕ d ϕ 2 = 0 (29.19) d s 2 = g t t d t 2 + 2 g t ϕ d t d ϕ + g ϕ ϕ d ϕ 2 = 0 {:(29.19)ds^(2)=g_(tt)dt^(2)+2g_(t phi)dtdphi+g_(phi phi)dphi^(2)=0:}\begin{equation*} \mathrm{d} s^{2}=g_{t t} \mathrm{~d} t^{2}+2 g_{t \phi} \mathrm{~d} t \mathrm{~d} \phi+g_{\phi \phi} \mathrm{d} \phi^{2}=0 \tag{29.19} \end{equation*}(29.19)ds2=gtt dt2+2gtϕ dt dϕ+gϕϕdϕ2=0
leading to an expression for the coordinate velocity in the ϕ ϕ phi\phiϕ direction of
(29.20) d ϕ d t = g t ϕ g ϕ ϕ ± [ ( g t ϕ g ϕ ϕ ) 2 g t t g ϕ ϕ ] 1 2 (29.20) d ϕ d t = g t ϕ g ϕ ϕ ± g t ϕ g ϕ ϕ 2 g t t g ϕ ϕ 1 2 {:(29.20)(dphi)/((d)t)=-(g_(t phi))/(g_(phi phi))+-[((g_(t phi))/(g_(phi phi)))^(2)-(g_(tt))/(g_(phi phi))]^((1)/(2)):}\begin{equation*} \frac{\mathrm{d} \phi}{\mathrm{~d} t}=-\frac{g_{t \phi}}{g_{\phi \phi}} \pm\left[\left(\frac{g_{t \phi}}{g_{\phi \phi}}\right)^{2}-\frac{g_{t t}}{g_{\phi \phi}}\right]^{\frac{1}{2}} \tag{29.20} \end{equation*}(29.20)dϕ dt=gtϕgϕϕ±[(gtϕgϕϕ)2gttgϕϕ]12
In the special case of the above example that 10 g t t = 0 10 g t t = 0 ^(10)g_(tt)=0{ }^{10} g_{t t}=010gtt=0, we have two solutions
(29.21) d ϕ d t = 2 g t ϕ g ϕ ϕ , d ϕ d t = 0 (29.21) d ϕ d t = 2 g t ϕ g ϕ ϕ , d ϕ d t = 0 {:(29.21)(dphi)/((d)t)=-(2g_(t phi))/(g_(phi phi))","quad(dphi)/((d)t)=0:}\begin{equation*} \frac{\mathrm{d} \phi}{\mathrm{~d} t}=-\frac{2 g_{t \phi}}{g_{\phi \phi}}, \quad \frac{\mathrm{d} \phi}{\mathrm{~d} t}=0 \tag{29.21} \end{equation*}(29.21)dϕ dt=2gtϕgϕϕ,dϕ dt=0
The first solution applies for photons sent in the same direction that the hole is spinning. The second applies for the photons sent against the direction of the hole's rotation. This last result is truly remarkable: in a region where g t t = 0 g t t = 0 g_(tt)=0g_{t t}=0gtt=0, a photon sent in opposite direction to that of the rotation of the mass initially cannot propagate. Since this applies for photons, which provide a universal speed limit via the light-cone structure, it applies for all particles, even those with large values of angular momentum. The effect can be further illustrated by considering a stationary observer, who has velocity with components u μ = ( u t , 0 , 0 , 0 ) u μ = u t , 0 , 0 , 0 u^(mu)=(u^(t),0,0,0)u^{\mu}=\left(u^{t}, 0,0,0\right)uμ=(ut,0,0,0). Since u 2 = 1 u 2 = 1 u^(2)=-1\boldsymbol{u}^{2}=-1u2=1 we must have g t t ( u t ) 2 = 1 g t t u t 2 = 1 g_(tt)(u^(t))^(2)=-1g_{t t}\left(u^{t}\right)^{2}=-1gtt(ut)2=1, which can't be satisfied if g t t g t t g_(tt)g_{t t}gtt vanishes, and so the observer cannot remain stationary and must start to rotate in the direction of the hole.
The surface on which g t t = 0 g t t = 0 g_(tt)=0g_{t t}=0gtt=0 lies outside the surface r + r + r_(+)r_{+}r+that defines the black hole's event horizon. It is called the stationary limit, since inside this surface it is impossible for particles to remain at rest. Setting g t t = 0 g t t = 0 g_(tt)=0g_{t t}=0gtt=0, we see that we obtain an ellipsoidal surface
(29.22) r 0 = M + ( M 2 a 2 cos 2 θ ) 1 2 (29.22) r 0 = M + M 2 a 2 cos 2 θ 1 2 {:(29.22)r_(0)=M+(M^(2)-a^(2)cos^(2)theta)^((1)/(2)):}\begin{equation*} r_{0}=M+\left(M^{2}-a^{2} \cos ^{2} \theta\right)^{\frac{1}{2}} \tag{29.22} \end{equation*}(29.22)r0=M+(M2a2cos2θ)12
Notice how the ellipsoid coincides with r + r + r_(+)r_{+}r+at the poles. The region r + r r 0 r + r r 0 r_(+) <= r <= r_(0)r_{+} \leq r \leq r_{0}r+rr0 is known as the ergosphere. 11 11 ^(11){ }^{11}11 It is part of region I and is shown in Figs. 29.3 and 29.5.
More insight into this spacetime can be found by further examining the behaviour of light in each of the regions.
10 10 ^(10){ }^{10}10 This occurs when ρ 2 = 2 M r ρ 2 = 2 M r rho^(2)=2Mr\rho^{2}=2 M rρ2=2Mr, resulting in a region of spacetime where this condition is met, as discussed below.
Fig. 29.5 The ergosphere for a nearextremal Kerr black hole, shown (a) from the side (i.e. observer at θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2, looking towards the origin); and (b) from the top (observer at θ = 0 θ = 0 theta=0\theta=0θ=0 ).
11 11 ^(11){ }^{11}11 Note that g t t < 0 g t t < 0 g_(tt) < 0g_{t t}<0gtt<0 outside the ergosphere, but g t t > 0 g t t > 0 g_(tt) > 0g_{t t}>0gtt>0 inside. This is the same as the state of affairs for the Schwarzschild metric either side of the Schwarzschild radius. We shall use this fact at the end of the chapter.

Example 29.6

The effective energy equation for photons orbiting in the plane θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 looks like
12 12 ^(12){ }^{12}12 See Exercise 29.7(b).
Fig. 29.6 The light-cone structure of the Kerr geometry.
(29.23) 1 L ~ 2 ( d r d λ ) 2 = 1 b 2 W eff ( r , b , σ ) (29.23) 1 L ~ 2 d r d λ 2 = 1 b 2 W eff ( r , b , σ ) {:(29.23)(1)/( tilde(L)^(2))(((d)r)/((d)lambda))^(2)=(1)/(b^(2))-W_(eff)(r","b","sigma):}\begin{equation*} \frac{1}{\tilde{L}^{2}}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}=\frac{1}{b^{2}}-W_{\mathrm{eff}}(r, b, \sigma) \tag{29.23} \end{equation*}(29.23)1L~2( dr dλ)2=1b2Weff(r,b,σ)
where, σ = sign ( L ~ ) σ = sign ( L ~ ) sigma=sign( tilde(L))\sigma=\operatorname{sign}(\tilde{L})σ=sign(L~) and 12 12 ^(12){ }^{12}12
(29.24) W eff ( r , b , σ ) = 1 r 2 [ 1 ( a b ) 2 2 M r ( 1 σ a b ) 2 ] . (29.24) W eff ( r , b , σ ) = 1 r 2 1 a b 2 2 M r 1 σ a b 2 . {:(29.24)W_(eff)(r","b","sigma)=(1)/(r^(2))[1-((a)/(b))^(2)-(2M)/(r)(1-sigma(a)/(b))^(2)].:}\begin{equation*} W_{\mathrm{eff}}(r, b, \sigma)=\frac{1}{r^{2}}\left[1-\left(\frac{a}{b}\right)^{2}-\frac{2 M}{r}\left(1-\sigma \frac{a}{b}\right)^{2}\right] . \tag{29.24} \end{equation*}(29.24)Weff(r,b,σ)=1r2[1(ab)22Mr(1σab)2].
The slightly peculiar factor of σ σ sigma\sigmaσ arises here telling us that the potential depends on the sign of the angular momentum of the photon. This is because the potential depends on whether the photon's angular momentum is in the direction of orbit of the black hole, or in the opposite direction. This is related to the frame-dragging that occurs for the Kerr black hole.
Finally, the light-cone structure of the black hole is shown in Fig. 29.6, where each of the regions can be identified and where the photon trajectories depend on the sense of rotation.

29.3 Interacting with the Kerr geometry

Roger Penrose suggested a method for extracting energy from a Kerr black hole. A particle starts at infinity and falls into the ergosphere of a black hole, where it decays into two particles. One of the particles falls through the event horizon and will eventually be consumed by the singularities; the other particle escapes. If the escaping particle carries away more energy to infinity than the original particle had, then we have removed energy from the black hole.
Locally, energy-momentum must be conserved, so we have that the incoming particle's 4 -momentum p in p in  p_("in ")\boldsymbol{p}_{\text {in }}pin  can be written as
(29.25) p in = p out + p bh (29.25) p in = p out + p bh {:(29.25)p_(in)=p_(out)+p_(bh):}\begin{equation*} \boldsymbol{p}_{\mathrm{in}}=\boldsymbol{p}_{\mathrm{out}}+\boldsymbol{p}_{\mathrm{bh}} \tag{29.25} \end{equation*}(29.25)pin=pout+pbh
where the outgoing particle has 4 -momentum p out p out  p_("out ")\boldsymbol{p}_{\text {out }}pout  and the particle that falls into the hole has momentum p bh p bh p_(bh)\boldsymbol{p}_{\mathrm{bh}}pbh.
As discussed in the last chapter, a particle's conserved energy, as measured by an observer at infinity, can be obtained from minus the dot product of the timelike Killing vector ξ ξ xi\boldsymbol{\xi}ξ and its momentum. As a result of a dot product of the last equation with ξ ξ xi\boldsymbol{\xi}ξ we find
(29.26) E out = E in E bh (29.26) E out = E in E bh {:(29.26)E_(out)^(oo)=E_(in)^(oo)-E_(bh)^(oo):}\begin{equation*} E_{\mathrm{out}}^{\infty}=E_{\mathrm{in}}^{\infty}-E_{\mathrm{bh}}^{\infty} \tag{29.26} \end{equation*}(29.26)Eout=EinEbh
So far so good: this is simply a form of conservation of energy which, under normal circumstances, implies that E out E in E out  E in E_("out ")^(oo) <= E_(in)^(oo)E_{\text {out }}^{\infty} \leq E_{\mathrm{in}}^{\infty}Eout Ein for positive E bh = p bh ξ E bh = p bh ξ E_(bh)^(oo)=-p_(bh)*xiE_{\mathrm{bh}}^{\infty}=-\boldsymbol{p}_{\mathrm{bh}} \cdot \boldsymbol{\xi}Ebh=pbhξ. The unusual feature comes from considering the special situation inside the ergosphere where the Killing vector ξ ξ xi\boldsymbol{\xi}ξ becomes
(29.27) ξ ξ = e t e t = g t t = ( 1 2 M r ρ 2 ) > 0 (29.27) ξ ξ = e t e t = g t t = 1 2 M r ρ 2 > 0 {:(29.27)xi*xi=e_(t)*e_(t)=g_(tt)=-(1-(2Mr)/(rho^(2))) > 0:}\begin{equation*} \boldsymbol{\xi} \cdot \boldsymbol{\xi}=\boldsymbol{e}_{t} \cdot \boldsymbol{e}_{t}=g_{t t}=-\left(1-\frac{2 M r}{\rho^{2}}\right)>0 \tag{29.27} \end{equation*}(29.27)ξξ=etet=gtt=(12Mrρ2)>0
and so the Killing vector is now spacelike. Just as in the argument for Hawking radiation in the last chapter, the dot product p bh ξ p bh ξ -p_(bh)*xi-\boldsymbol{p}_{\mathrm{bh}} \cdot \boldsymbol{\xi}pbhξ is now no longer constrained to be positive. Therefore, if the decay has E bh = p bh ξ < 0 E bh = p bh ξ < 0 E_(bh)^(oo)=-p_(bh)*xi < 0E_{\mathrm{bh}}^{\infty}=-\boldsymbol{p}_{\mathrm{bh}} \cdot \boldsymbol{\xi}<0Ebh=pbhξ<0 we have have E out > E in E out  > E in E_("out ")^(oo) > E_(in)^(oo)E_{\text {out }}^{\infty}>E_{\mathrm{in}}^{\infty}Eout >Ein and energy can be extracted from the black hole. The consequence is that negative E bh E bh E_(bh)E_{\mathrm{bh}}Ebh will consume some of the mass of the black hole. It also removes some angular momentum, as demonstrated in the next example.

Example 29.7

Consider a rotating observer who remains at fixed r r rrr and θ θ theta\thetaθ in a Kerr black hole. (They cannot help but be rotated in ϕ ϕ phi\phiϕ.) They have a velocity
(29.28) u μ = u t ( 1 , 0 , 0 , Ω ) (29.28) u μ = u t ( 1 , 0 , 0 , Ω ) {:(29.28)u^(mu)=u^(t)(1","0","0","Omega):}\begin{equation*} u^{\mu}=u^{t}(1,0,0, \Omega) \tag{29.28} \end{equation*}(29.28)uμ=ut(1,0,0,Ω)
where Ω Ω Omega\OmegaΩ is the angular speed. This velocity can be written in vector form using the Killing vectors as
(29.29) u obs = u t ( ξ + Ω η ) (29.29) u obs = u t ( ξ + Ω η ) {:(29.29)u_(obs)=u^(t)(xi+Omega eta):}\begin{equation*} \boldsymbol{u}_{\mathrm{obs}}=u^{t}(\boldsymbol{\xi}+\Omega \boldsymbol{\eta}) \tag{29.29} \end{equation*}(29.29)uobs=ut(ξ+Ωη)
The observer measures a positive energy for the particle, and so using E = u obs p E = u obs p E=-u_(obs)*pE=-\boldsymbol{u}_{\mathrm{obs}} \cdot \boldsymbol{p}E=uobsp, we find
(29.30) ( ξ + Ω η ) p bh 0 (29.30) ( ξ + Ω η ) p bh 0 {:(29.30)-(xi+Omega eta)*p_(bh) >= 0:}\begin{equation*} -(\boldsymbol{\xi}+\Omega \boldsymbol{\eta}) \cdot \boldsymbol{p}_{\mathrm{bh}} \geq 0 \tag{29.30} \end{equation*}(29.30)(ξ+Ωη)pbh0
for any value of Ω Ω Omega\OmegaΩ that allows u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1. Using E = ξ p E = ξ p E=-xi*pE=-\boldsymbol{\xi} \cdot \boldsymbol{p}E=ξp and L = η p L = η p L=eta*pL=\boldsymbol{\eta} \cdot \boldsymbol{p}L=ηp, we have
(29.31) E bh Ω L bh (29.31) E bh Ω L bh {:(29.31)E_(bh) >= OmegaL_(bh):}\begin{equation*} E_{\mathrm{bh}} \geq \Omega L_{\mathrm{bh}} \tag{29.31} \end{equation*}(29.31)EbhΩLbh
where L bh L bh L_(bh)L_{\mathrm{bh}}Lbh is the angular momentum of the particle that falls into the black hole. Since Ω Ω Omega\OmegaΩ is positive (since the observer must rotate in the same direction as the black hole), then L bh L bh L_(bh)L_{\mathrm{bh}}Lbh must be negative if we are to have E bf < 0 E bf < 0 E_(bf) < 0E_{\mathrm{bf}}<0Ebf<0 and extract energy from the hole. As a result, the angular momentum of the hole is reduced.

Chapter summary

  • Charged black holes are unlikely to be realized in Nature. They are described by the Reissner-Nordstöm coordinates and feature three different regions of spacetime.
  • The singularity at the centre of a charged black hole is repulsive. A traveller can use the wormhole created by the charge to tunnel into a new Universe.
  • Rotating black holes are described by the Kerr geometry given in Boyer-Lindquist coordinates. There is a ring of physical singularities and an ellipsoidal event horizon at r 0 r 0 r_(0)r_{0}r0.
  • In the Kerr geometry, the region between the coordinate singularity at r + r + r_(+)r_{+}r+and the surface r 0 r 0 r_(0)r_{0}r0 where g t t g t t g_(tt)g_{t t}gtt vanishes is the ergosphere, in which it is impossible for a particle to remain stationary.
The reader will by now have built up a healthy respect for the Kerr solution. However, like many results in mathematical physics, it's rather hard to imagine how anyone could have come up with it! Kerr's announcement of the solution in 1963 did not give a derivation. A paper by Kerr and Schild from 1965 does so, but warns "the calculations giving these results are by no means simple", while Landau and Lifshitz (1975) say that "there is no constructive analytic derivation of the metric that is adequate in its physical ideas, and even a check of this solution of Einstein's equations involves cumbersome calculations". However, later work has elucidated the physical motivation, with Chandrasekhar describing the derivation he gives in his (excellent, but very advanced) book as "really very simple" with an "adequate base of physical and mathematical motivations". It is perhaps sobering that even Chandrasekhar provides a warning regarding perturbations to the Kerr solution, saying "the reductions that are necessary to go from one step to another are often very elaborate and, on occasion, may require as many as ten, twenty, or even fifty pages". He goes on to say that a copy of his derivations "(in some 600 legal-size pages and six additional notebooks)" are available via the Joseph Regenstein Library, University of Chicago, "in case some reader may wish to undertake a careful scrutiny of the entire development".

Exercises

(29.1) Verify that the coordinate singularities for the (29.5) By evaluating the determinant g g ggg and integrating, Reissner-Nordström line element occur at r ± = r ± = r_(+-)=r_{ \pm}=r±= M ± ( M 2 q 2 ) 1 2 M ± M 2 q 2 1 2 M+-(M^(2)-q^(2))^((1)/(2))M \pm\left(M^{2}-q^{2}\right)^{\frac{1}{2}}M±(M2q2)12.
(29.2) (a) Show that g t t g t t g_(tt)g_{t t}gtt for the Kerr solution can also be
rewritten as (29.32) g t t = Δ a 2 sin 2 θ ρ 2  rewritten as  (29.32) g t t = Δ a 2 sin 2 θ ρ 2 {:[" rewritten as "],[(29.32)qquadg_(tt)=-(Delta-a^(2)sin^(2)theta)/(rho^(2))]:}\begin{align*} & \text { rewritten as } \\ & \qquad g_{t t}=-\frac{\Delta-a^{2} \sin ^{2} \theta}{\rho^{2}} \tag{29.32} \end{align*} rewritten as (29.32)gtt=Δa2sin2θρ2
(b) Show further that g ϕ ϕ can be rewritten as g ϕ ϕ = ( 2 M r a 2 sin 2 θ ρ 2 + r 2 + a 2 ) sin 2 θ .  (b) Show further that  g ϕ ϕ  can be rewritten as  g ϕ ϕ = 2 M r a 2 sin 2 θ ρ 2 + r 2 + a 2 sin 2 θ . {:[" (b) Show further that "g_(phi phi)" can be rewritten as "],[g_(phi phi)=((2Mra^(2)sin^(2)theta)/(rho^(2))+r^(2)+a^(2))sin^(2)theta.]:}\begin{aligned} & \text { (b) Show further that } g_{\phi \phi} \text { can be rewritten as } \\ & g_{\phi \phi}=\left(\frac{2 M r a^{2} \sin ^{2} \theta}{\rho^{2}}+r^{2}+a^{2}\right) \sin ^{2} \theta . \end{aligned} (b) Show further that gϕϕ can be rewritten as gϕϕ=(2Mra2sin2θρ2+r2+a2)sin2θ.
(29.3) Verify eqn 29.6 .
(29.4) Show that
d s 2 = ρ + d θ 2 + ( r + 2 + a 2 + 2 M r + a 2 sin 2 θ ρ + 2 ) sin 2 θ d ϕ 2 (29.34) = ρ + 2 d θ 2 + ( 2 M r + ρ + ) 2 sin 2 θ d ϕ 2 . d s 2 = ρ + d θ 2 + r + 2 + a 2 + 2 M r + a 2 sin 2 θ ρ + 2 sin 2 θ d ϕ 2 (29.34) = ρ + 2 d θ 2 + 2 M r + ρ + 2 sin 2 θ d ϕ 2 . {:[ds^(2)=rho_(+)dtheta^(2)],[+(r_(+)^(2)+a^(2)+(2Mr_(+)a^(2)sin^(2)theta)/(rho_(+)^(2)))sin^(2)thetadphi^(2)],[(29.34)=rho_(+)^(2)dtheta^(2)+((2Mr_(+))/(rho_(+)))^(2)sin^(2)thetadphi^(2).]:}\begin{align*} \mathrm{d} s^{2}= & \rho_{+} \mathrm{d} \theta^{2} \\ & +\left(r_{+}^{2}+a^{2}+\frac{2 M r_{+} a^{2} \sin ^{2} \theta}{\rho_{+}^{2}}\right) \sin ^{2} \theta \mathrm{~d} \phi^{2} \\ = & \rho_{+}^{2} \mathrm{~d} \theta^{2}+\left(\frac{2 M r_{+}}{\rho_{+}}\right)^{2} \sin ^{2} \theta \mathrm{~d} \phi^{2} . \tag{29.34} \end{align*}ds2=ρ+dθ2+(r+2+a2+2Mr+a2sin2θρ+2)sin2θ dϕ2(29.34)=ρ+2 dθ2+(2Mr+ρ+)2sin2θ dϕ2.
A = 0 2 π d ϕ 0 π d θ g = 8 π M r + (29.35) (29.6) Verify eqns 29.12 by inverting the matrix A = 0 2 π d ϕ 0 π d θ g = 8 π M r + (29.35) (29.6) Verify eqns  29.12  by inverting the matrix  {:[qquad A=int_(0)^(2pi)dphiint_(0)^(pi)dtheta g=8pi Mr_(+)],[(29.35)(29.6) Verify eqns 29.12" by inverting the matrix "]:}\begin{gather*} \qquad A=\int_{0}^{2 \pi} \mathrm{~d} \phi \int_{0}^{\pi} \mathrm{d} \theta g=8 \pi M r_{+} \\ \text {(29.6) Verify eqns } 29.12 \text { by inverting the matrix } \tag{29.35} \end{gather*}A=02π dϕ0πdθg=8πMr+(29.35)(29.6) Verify eqns 29.12 by inverting the matrix 
(29.36) ( g t t g t ϕ g ϕ t g ϕ ϕ ) . (29.36) g t t g t ϕ g ϕ t g ϕ ϕ . {:(29.36)([g_(tt),g_(t phi)],[g_(phi t),g_(phi phi)]).:}\left(\begin{array}{ll} g_{t t} & g_{t \phi} \tag{29.36}\\ g_{\phi t} & g_{\phi \phi} \end{array}\right) .(29.36)(gttgtϕgϕtgϕϕ). show that
(29.7) (a) Verify eqn 29.18 .
(b) Verify eqn 29.24 .
(29.8) What is the interval in proper time required to orbit a black hole described by the Kerr metric at fixed θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2, constant r = R r = R r=Rr=Rr=R and coordinate speed v = r d ϕ / d t v = r d ϕ / d t v=rdphi//dtv=r \mathrm{~d} \phi / \mathrm{d} tv=r dϕ/dt.
Hint: An analogous problem appears in Chapter 23 in Exercise 23.1.

Part V

Geometry

In Part V of the book, we take a closer look at the mathematics of differential geometry and curvature. These ideas will be put to work in the final part of the book.
  • In Chapter 30, we look at how curved lines and surfaces can be described mathematically using classical differential calculus.
  • Then, in Chapters 31 and 32 , we proceed to upgrade our basic mathematical machinery to provide a modern treatment of tensors.
  • In Chapters 33 and 34, we meet the derivatives that are naturally adapted to a coordinate-free description of a manifold.
  • In Chapter 35, we once again encounter curvature, but now from a thoroughly modern point of view. This enables us, in Chapter 36, to compute the curvature from a metric using an incredibly efficient procedure invented by Élie Cartan.
  • In Chapter 37, we discuss how to deal with areas and volumes in coordinate-free language.
  • In Chapter 38, we see how this new approach gives insights into how we understand integrals in physics.
  • Appendix C contains some further description of the mathematical background behind this part of the book and can be consulted for interest or reference.
For many readers with a physics background, the material in this part of the book might seem rather unfamiliar. It's worth keeping in mind that Einstein was also initially unfamiliar with differential geometry when he started working on general relativity. He benefitted from the help of his friend Marcel Grossmann (1878-1936) for an introduction to many of the ideas.

30

30.1 Curvature of a line
30.4 Gauss' equation 314
30.5 Intrinsic and extrinsic curvature
30.6 Riemann's project 318
Chapter summary 320
Exercises
\curvearrowright We call the approach outlined in this chapter classical curvature to distinguish it from the more modern, coordinate-free approach outlined in the following chapters. As a result, this chapter can be skipped if so desired.
Fig. 30.1 The outline of this guitar traces out a path in space. The curvature of the upper region with radius ρ 1 ρ 1 rho_(1)\rho_{1}ρ1 is greater than that with radius ρ 2 ρ 2 rho_(2)\rho_{2}ρ2.
1 1 ^(1){ }^{1}1 Augustin-Louis Cauchy (1789-1857).

Classical curvature

This paper will no doubt be found interesting by those who take an interest in it.
John Dalton (1766-1844)
The study of curvature has a rich history in mathematics. Our knowledge of the curvature of a line was developed by applying calculus to geometry, thereby founding the study of differential geometry. In this chapter, we follow some of the key developments made by mathematicians such as Leonhard Euler and Augustin-Louis Cauchy that culminate in the Frenet-Serret equations, which represent an important milestone in our understanding of the curvature of paths in space. Our knowledge of the curvature of planes is largely based around the work of two mathematicians: Carl Friedrich Gauss and Bernhard Riemann. Gauss showed how the curvature of a two-dimensional plane can be described by measurements made by beings entirely confined to that plane. Riemann generalized the result to higher dimensions, and it is Riemann's toolkit of insights that are incorporated into general relativity.

30.1 Curvature of a line

How do we measure the curvature of a path in two dimensions? The curved edge of the object shown in Fig. 30.1 has lots of features, but it's reasonable to say that the part at the top has a larger curvature than the part at the bottom, since its tangent vectors turn more rapidly as we traverse that part of the curve. If this isn't obvious, imagine driving in a car traversing the two indicated bends of the curve at a constant speed and consider how much you would need to turn the steering wheel in each case.
In order to quantify curvature, let's identify a figure, such as the circle, against which curvature can be compared. The circle, after all, involves a line element constantly changing its direction about the central point, so seems to have a suitably curvy quality. A small circle has greater curvature than a circle of larger radius, while a straight line has no curvature. This suggests we can use the reciprocal of a circle's radius as the measure of curvature. Let's see how we might do this systematically, by following Augustin-Louis Cauchy, 1 1 ^(1){ }^{1}1 and evaluate the curvature at a specific point P P P\mathcal{P}P on the curve.
A circle's centre can always be found by seeing where two normals to the circle intersect. We will use this fact to identify the osculating circle, 2 2 ^(2){ }^{2}2 which will provide our measure of curvature. Referring to Fig. 30.2 where we determine the curvature of the curve at point P P P\mathcal{P}P. We define the centre of curvature C C C\mathcal{C}C as the point where two closely spaced normals to the curve intersect, allowing us to draw the osculating circle, centred at C C C\mathcal{C}C and intersecting P P P\mathcal{P}P. The radius of curvature ρ ρ rho\rhoρ is then the distance from P P P\mathcal{P}P to C C C\mathcal{C}C and we define the curvature κ κ kappa\kappaκ to be κ = 1 / ρ κ = 1 / ρ kappa=1//rho\kappa=1 / \rhoκ=1/ρ. Referring back to Fig. 30.1, we see how this does the job: the feature at the bottom has an osculating circle with a smaller radius and, therefore, a larger curvature κ κ kappa\kappaκ, as we had suggested. The curvature of a curve at a point P P P\mathcal{P}P may, therefore, be thought of as the inverse radius of a circle that locally follows the curve at P P P\mathcal{P}P. We could repeat this procedure as we move along the curve, assessing the curvature at different points.
Example 30.1
Recall that a curve is parametrized by some parameter λ λ lambda\lambdaλ that varies along the curve. We shall choose to use length parametrization 3 3 ^(3){ }^{3}3 so that we set λ λ lambda\lambdaλ equal to the distance s s sss measured along the curve (Fig. 30.3). An alternative way to find the curvature is to see how the directions of the normals, or equivalently, 4 4 ^(4){ }^{4}4 the tangent vectors to the curve, rotate as we traverse the curve. Specifically, we work out the tangent vectors at two ends of a short section of curve of length d s d s ds\mathrm{d} sds and measure the angle d θ d θ dtheta\mathrm{d} \thetadθ between the two tangents. Extracting the tangent vectors from the figure, normalizing them, and arranging them in the unit circle, as shown in Fig. 30.4, provides us with a method for calculating the curvature
(30.1) κ ( s ) = lim Δ s 0 ( Angle Δ θ swept out on the unit circle by normals or tangents ) ( Corresponding arc length Δ s on actual curve ) . (30.1) κ ( s ) = lim Δ s 0 (  Angle  Δ θ  swept out on   the unit circle by normals or tangents  ) (  Corresponding arc length  Δ s  on actual curve  ) . {:(30.1)kappa(s)=lim_(Delta s rarr0)(((" Angle "Delta theta" swept out on ")/(" the unit circle by normals or tangents ")))/(((" Corresponding arc length "Delta s)/(" on actual curve "))).:}\begin{equation*} \kappa(s)=\lim _{\Delta s \rightarrow 0} \frac{\binom{\text { Angle } \Delta \theta \text { swept out on }}{\text { the unit circle by normals or tangents }}}{\binom{\text { Corresponding arc length } \Delta s}{\text { on actual curve }}} . \tag{30.1} \end{equation*}(30.1)κ(s)=limΔs0( Angle Δθ swept out on  the unit circle by normals or tangents )( Corresponding arc length Δs on actual curve ).
Putting this together we have
(30.2) κ ( s ) = 1 ρ ( s ) = d θ ( s ) d s (30.2) κ ( s ) = 1 ρ ( s ) = d θ ( s ) d s {:(30.2)kappa(s)=(1)/(rho(s))=(dtheta(s))/(ds):}\begin{equation*} \kappa(s)=\frac{1}{\rho(s)}=\frac{\mathrm{d} \theta(s)}{\mathrm{d} s} \tag{30.2} \end{equation*}(30.2)κ(s)=1ρ(s)=dθ(s)ds
To see that this makes sense, consider a circle of radius ρ ρ rho\rhoρ shown in Fig. 30.4. The tangents (and normals) sweep out angle Δ θ Δ θ Delta theta\Delta \thetaΔθ corresponding to the arc length ρ Δ θ ρ Δ θ rho Delta theta\rho \Delta \thetaρΔθ. It's fairly easy to see that the curvature is κ = 1 / ρ κ = 1 / ρ kappa=1//rho\kappa=1 / \rhoκ=1/ρ.
A slightly more coordinate-friendly method of assessing the curvature of a line is to measure the departure y y yyy of the curve from a tangent line as a function of x x xxx. The curvature κ κ kappa\kappaκ is then written as
(30.3) κ = 1 ρ = d 2 y d x 2 (30.3) κ = 1 ρ = d 2 y d x 2 {:(30.3)kappa=(1)/(rho)=(d^(2)y)/((d)x^(2)):}\begin{equation*} \kappa=\frac{1}{\rho}=\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} \tag{30.3} \end{equation*}(30.3)κ=1ρ=d2y dx2
Example 30.2
How does this latter method work? Consider the tangent to the curve at point P P P\mathcal{P}P with coordinate x p x p x_(p)x_{p}xp. If x x xxx is the coordinate measured along the tangent relative to x p x p x_(p)x_{p}xp, then y ( x ) y ( x ) y(x)y(x)y(x) is the height of the curve above the tangent. For small x x xxx we can expand
(30.4) y ( x ) = y ( x p ) + d y d x | x p ( x x p ) + d 2 y d x 2 | x p ( x x p ) 2 2 + (30.4) y ( x ) = y x p + d y d x x p x x p + d 2 y d x 2 x p x x p 2 2 + {:(30.4)y(x)=y(x_(p))+(dy)/((d)x)|_(x_(p))(x-x_(p))+(d^(2)y)/((d)x^(2))|_(x_(p))((x-x_(p))^(2))/(2)+dots:}\begin{equation*} y(x)=y\left(x_{p}\right)+\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{x_{p}}\left(x-x_{p}\right)+\left.\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right|_{x_{p}} \frac{\left(x-x_{p}\right)^{2}}{2}+\ldots \tag{30.4} \end{equation*}(30.4)y(x)=y(xp)+dy dx|xp(xxp)+d2y dx2|xp(xxp)22+
2 2 ^(2){ }^{2}2 Outside of mathematics, osculate is a verb derived from the Latin verb osculare, meaning to kiss (Da mihi osculum -=\equiv give me a kiss).
Fig. 30.2 The osculating circle for a point P P P\mathcal{P}P on a curve. It is centred on C C C\mathcal{C}C and has radius ρ ρ rho\rhoρ.
Fig. 30.3 Length parametrization of the curve uses the distance s s sss along the curve as a parameter.
3 3 ^(3){ }^{3}3 See Chapter 9
4 4 ^(4){ }^{4}4 The normals to a curve are simply the tangents rotated by 90 90 90^(@)90^{\circ}90.
Fig. 30.4 Top: a curve with unit tangent vectors evolving over a distance Δ s Δ s Delta s\Delta sΔs. Bottom: arranging the tangent vectors in the unit circle provides a measure of Δ θ Δ θ Delta theta\Delta \thetaΔθ.
5 5 ^(5){ }^{5}5 Notice that, because we measure cur vature in terms of infinitesimals, in or der to extract the curvature κ κ kappa\kappaκ in the coordinate version we compare the curvature to a parabola (i.e. the height of the circle close to x p x p x_(p)x_{p}xp is y = x 2 / 2 ρ y = x 2 / 2 ρ y=x^(2)//2rhoy=x^{2} / 2 \rhoy=x2/2ρ ) which has d 2 y / d x 2 = d 2 y / d x 2 = d^(2)y//dx^(2)=\mathrm{d}^{2} y / \mathrm{d} x^{2}=d2y/dx2= (const.). We shall re turn to this when we look at the curvature of two-dimensional surfaces.
6 6 ^(6){ }^{6}6 This is a temporary measure for this chapter.
7 7 ^(7){ }^{7}7 This use of the displacement X X X\boldsymbol{X}X relies on the one-dimensional curve being embedded in flat three-dimensional space We shall see in the next chapter that it is not a strategy that we shall attempt in curved spacetime.
Fig. 30.5 The triplet formed from tan gent t t t\boldsymbol{t}t, normal p p p\boldsymbol{p}p and binormal b = b = b=\boldsymbol{b}=b= t × p t × p t xx p\boldsymbol{t} \times \boldsymbol{p}t×p. The osculating plane is defined by t t t\boldsymbol{t}t and p p p\boldsymbol{p}p.
As the tangent is defined by d y / d x = 0 d y / d x = 0 dy//dx=0\mathrm{d} y / \mathrm{d} x=0dy/dx=0, taking x p = 0 x p = 0 x_(p)=0x_{p}=0xp=0 we have y ( x ) = κ x 2 / 2 y ( x ) = κ x 2 / 2 y(x)=kappax^(2)//2y(x)=\kappa x^{2} / 2y(x)=κx2/2.
Next, imagine a circle of radius ρ ρ rho\rhoρ which is tangent to the curve at p p ppp. Its height is y = ρ ( 1 cos θ ) y = ρ ( 1 cos θ ) y=rho(1-cos theta)y=\rho(1-\cos \theta)y=ρ(1cosθ). For small θ θ theta\thetaθ, we have θ = x / ρ θ = x / ρ theta=x//rho\theta=x / \rhoθ=x/ρ and so y = x 2 / 2 ρ y = x 2 / 2 ρ y=x^(2)//2rhoy=x^{2} / 2 \rhoy=x2/2ρ and we conclude that κ = 1 / ρ κ = 1 / ρ kappa=1//rho\kappa=1 / \rhoκ=1/ρ as we had before. 5 5 ^(5){ }^{5}5

30.2 Curvature with vectors

A still more formal way of examining the curvature properties of a line is to invoke differential geometry. We will work in flat three-dimensional space, denoting 3 -vectors by bold letters. 6 6 ^(6){ }^{6}6 As usual, we parametrize a curve using the distance parameter s s sss. We say that the curve is defined by vectors X ( s ) X ( s ) X(s)\boldsymbol{X}(s)X(s) (i.e. insert a value of s s sss giving the position on the curve and return a vector taking us from the origin to that point on the curve). 7 7 ^(7){ }^{7}7 This curve has a tangent
(30.5) t = d X d s (30.5) t = d X d s {:(30.5)t=(dX)/((d)s):}\begin{equation*} \boldsymbol{t}=\frac{\mathrm{d} \boldsymbol{X}}{\mathrm{~d} s} \tag{30.5} \end{equation*}(30.5)t=dX ds
The tangent vector t t t\boldsymbol{t}t can be expanded in terms of coordinates x μ x μ x^(mu)x^{\mu}xμ by using basis vectors defined along the curve and writing t = t μ e μ t = t μ e μ t=t^(mu)e_(mu)t=t^{\mu} e_{\mu}t=tμeμ. Then we have
(30.6) t = d X d s = d x μ d s X x μ t μ e μ . (30.6) t = d X d s = d x μ d s X x μ t μ e μ . {:(30.6)t=(dX)/((d)s)=(dx^(mu))/(ds)(del X)/(delx^(mu))-=t^(mu)e_(mu).:}\begin{equation*} \boldsymbol{t}=\frac{\mathrm{d} \boldsymbol{X}}{\mathrm{~d} s}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} s} \frac{\partial \boldsymbol{X}}{\partial x^{\mu}} \equiv t^{\mu} \boldsymbol{e}_{\mu} . \tag{30.6} \end{equation*}(30.6)t=dX ds=dxμdsXxμtμeμ.
The tangent vector's components are t μ = d μ μ d s t μ = d μ μ d s t^(mu)=(dmu^(mu))/(ds)t^{\mu}=\frac{\mathrm{d} \mu^{\mu}}{\mathrm{d} s}tμ=dμμds and the basis vectors on the curve are e μ = X x μ e μ = X x μ e_(mu)=(del X)/(delx^(mu))\boldsymbol{e}_{\mu}=\frac{\partial \boldsymbol{X}}{\partial x^{\mu}}eμ=Xxμ.
If s s sss parametrizes length along the curve then d s 2 = d X d X d s 2 = d X d X ds^(2)=dX*dX\mathrm{d} s^{2}=\mathrm{d} \boldsymbol{X} \cdot \mathrm{d} \boldsymbol{X}ds2=dXdX, and so we must have t t = 1 t t = 1 t*t=1\boldsymbol{t} \cdot \boldsymbol{t}=1tt=1, which is to say that the tangents are normalized to unity. Differentiation of t t = 1 implies t t = 1 implies t*t=1implies\boldsymbol{t} \cdot \boldsymbol{t}=1 \mathrm{implies}tt=1implies that t d t d s = 0 t d t d s = 0 t*(dt)/(ds)=0\boldsymbol{t} \cdot \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{d} s}=0tdtds=0. We therefore define
(30.7) d t d s = κ p (30.7) d t d s = κ p {:(30.7)(dt)/((d)s)=kappa p:}\begin{equation*} \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{~d} s}=\kappa \boldsymbol{p} \tag{30.7} \end{equation*}(30.7)dt ds=κp
where p p p\boldsymbol{p}p, known as the principal normal vector, is orthogonal to t t t\boldsymbol{t}t, and we also fix p p = 1 p p = 1 p*p=1\boldsymbol{p} \cdot \boldsymbol{p}=1pp=1. As we prove in the next example, the constant κ κ kappa\kappaκ is the curvature of the curve that we defined above. The vectors t t ttt and p p p\boldsymbol{p}p span a two-dimensional surface known as the osculating plane, as shown in Fig. 30.5. The osculating plane is carried along as we traverse the curve. If the curve is confined to a plane (a so-called plane curve), the osculating plane will always be the same plane, although the directions of p p p\boldsymbol{p}p and t t t\boldsymbol{t}t will change.
Example 30.3
We pause to check that κ κ kappa\kappaκ in the previous equation is indeed the curvature of the curve. Let Δ θ Δ θ Delta theta\Delta \thetaΔθ be the angle between tangent t ( s ) t ( s ) t(s)\boldsymbol{t}(s)t(s) at X ( s ) X ( s ) X(s)\boldsymbol{X}(s)X(s) and tangent t ( s + Δ s ) t ( s + Δ s ) t(s+Delta s)\boldsymbol{t}(s+\Delta s)t(s+Δs) at the point X ( s + Δ s ) X ( s + Δ s ) X(s+Delta s)\boldsymbol{X}(s+\Delta s)X(s+Δs). If the tangent is normalized to unity, the vector | t ( s + Δ s ) t ( s ) | | t ( s + Δ s ) t ( s ) | |t(s+Delta s)-t(s)||\boldsymbol{t}(s+\Delta s)-\boldsymbol{t}(s)||t(s+Δs)t(s)| is the base of an isosceles triangle with unit sides. Hence,
(30.8) | t ( s + Δ s ) t ( s ) | = 2 sin ( Δ θ 2 ) = Δ θ + O ( Δ θ 3 ) (30.8) | t ( s + Δ s ) t ( s ) | = 2 sin Δ θ 2 = Δ θ + O Δ θ 3 {:(30.8)|t(s+Delta s)-t(s)|=2sin((Delta theta)/(2))=Delta theta+O(Deltatheta^(3)):}\begin{equation*} |t(s+\Delta s)-t(s)|=2 \sin \left(\frac{\Delta \theta}{2}\right)=\Delta \theta+O\left(\Delta \theta^{3}\right) \tag{30.8} \end{equation*}(30.8)|t(s+Δs)t(s)|=2sin(Δθ2)=Δθ+O(Δθ3)
Then, noting that | p | = 1 | p | = 1 |p|=1|\boldsymbol{p}|=1|p|=1 and writing 8 t ˙ = d t / d s 8 t ˙ = d t / d s ^(8)t^(˙)=dt//ds{ }^{8} \dot{\boldsymbol{t}}=\mathrm{d} \boldsymbol{t} / \mathrm{d} s8t˙=dt/ds, we have
| κ | = | t ˙ | = | lim Δ s 0 t ( s + Δ s ) t ( s ) Δ s | (30.9) = lim Δ s 0 Δ θ + O ( Δ θ 3 ) Δ s | κ | = | t ˙ | = lim Δ s 0 t ( s + Δ s ) t ( s ) Δ s (30.9) = lim Δ s 0 Δ θ + O Δ θ 3 Δ s {:[|kappa|=|t^(˙)|=|lim_(Delta s rarr0)(t(s+Delta s)-t(s))/(Delta s)|],[(30.9)=lim_(Delta s rarr0)(Delta theta+O(Deltatheta^(3)))/(Delta s)]:}\begin{align*} |\kappa|=|\dot{\boldsymbol{t}}| & =\left|\lim _{\Delta s \rightarrow 0} \frac{\boldsymbol{t}(s+\Delta s)-\boldsymbol{t}(s)}{\Delta s}\right| \\ & =\lim _{\Delta s \rightarrow 0} \frac{\Delta \theta+O\left(\Delta \theta^{3}\right)}{\Delta s} \tag{30.9} \end{align*}|κ|=|t˙|=|limΔs0t(s+Δs)t(s)Δs|(30.9)=limΔs0Δθ+O(Δθ3)Δs
This leads to
(30.10) | κ | = lim Δ s 0 Δ θ Δ s = d θ d s (30.10) | κ | = lim Δ s 0 Δ θ Δ s = d θ d s {:(30.10)|kappa|=lim_(Delta s rarr0)(Delta theta)/(Delta s)=(dtheta)/((d)s):}\begin{equation*} |\kappa|=\lim _{\Delta s \rightarrow 0} \frac{\Delta \theta}{\Delta s}=\frac{\mathrm{d} \theta}{\mathrm{~d} s} \tag{30.10} \end{equation*}(30.10)|κ|=limΔs0ΔθΔs=dθ ds
demonstrating that κ κ kappa\kappaκ is the curvature of the curve, just as we had before.
To describe curves in three dimensions (i.e. curves that are not plane curves) it would be useful to have an orthogonal triplet of vectors that spans the space. To make such a triplet, we also define the binormal vector, another unit vector ( b b = 1 ) ( b b = 1 ) (b*b=1)(\boldsymbol{b} \cdot \boldsymbol{b}=1)(bb=1), as
(30.12) b ( s ) = t ( s ) × p ( s ) (30.12) b ( s ) = t ( s ) × p ( s ) {:(30.12)b(s)=t(s)xx p(s):}\begin{equation*} \boldsymbol{b}(s)=\boldsymbol{t}(s) \times \boldsymbol{p}(s) \tag{30.12} \end{equation*}(30.12)b(s)=t(s)×p(s)
Example 30.4
This definition fixes the direction of b ˙ d b / d b ˙ d b / d b^(˙)-=db//d\dot{\boldsymbol{b}} \equiv \mathrm{d} \boldsymbol{b} / \mathrm{d}b˙db/d s. Since we define b b = 1 b b = 1 b*b=1\boldsymbol{b} \cdot \boldsymbol{b}=1bb=1, we must have b b ˙ = 0 b b ˙ = 0 b*b^(˙)=0\boldsymbol{b} \cdot \dot{\boldsymbol{b}}=0bb˙=0. Noting also that b t = 0 b t = 0 b*t=0\boldsymbol{b} \cdot \boldsymbol{t}=0bt=0, we have
b ˙ t + b t ˙ = 0 b ˙ t = b t ˙ b ˙ t + b t ˙ = 0 b ˙ t = b t ˙ {:[b^(˙)*t+b*t^(˙)=0],[b^(˙)*t=-b*t^(˙)]:}\begin{aligned} \dot{\boldsymbol{b}} \cdot \boldsymbol{t}+\boldsymbol{b} \cdot \dot{\boldsymbol{t}} & =0 \\ \dot{\boldsymbol{b}} \cdot \boldsymbol{t} & =-\boldsymbol{b} \cdot \dot{\boldsymbol{t}} \end{aligned}b˙t+bt˙=0b˙t=bt˙
(30.13) = κ b p = 0 (30.13) = κ b p = 0 {:(30.13)=-kappa b*p=0:}\begin{equation*} =-\kappa \boldsymbol{b} \cdot \boldsymbol{p}=0 \tag{30.13} \end{equation*}(30.13)=κbp=0
We conclude that b ˙ b ˙ b^(˙)\dot{\boldsymbol{b}}b˙ is perpendicular to b b b\boldsymbol{b}b and t t t\boldsymbol{t}t, so must lie along p p p\boldsymbol{p}p.
Motivated by the previous example, we define
(30.14) d b d s = τ p (30.14) d b d s = τ p {:(30.14)(db)/((d)s)=-tau p:}\begin{equation*} \frac{\mathrm{d} b}{\mathrm{~d} s}=-\tau \boldsymbol{p} \tag{30.14} \end{equation*}(30.14)db ds=τp
where τ τ tau\tauτ is known as the torsion of the curve. Since the three vectors t , p t , p t,p\boldsymbol{t}, \boldsymbol{p}t,p and b b b\boldsymbol{b}b form an orthogonal triplet, we can also write p = b × t p = b × t p=b xx t\boldsymbol{p}=\boldsymbol{b} \times \boldsymbol{t}p=b×t. Differentiating this latter equation, we find that
(30.15) d p d s = d b d s × t + b × d t d s = τ b κ t (30.15) d p d s = d b d s × t + b × d t d s = τ b κ t {:(30.15)(dp)/((d)s)=(db)/((d)s)xx t+b xx(dt)/((d)s)=tau b-kappa t:}\begin{equation*} \frac{\mathrm{d} \boldsymbol{p}}{\mathrm{~d} s}=\frac{\mathrm{d} \boldsymbol{b}}{\mathrm{~d} s} \times \boldsymbol{t}+\boldsymbol{b} \times \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{~d} s}=\tau \boldsymbol{b}-\kappa \boldsymbol{t} \tag{30.15} \end{equation*}(30.15)dp ds=db ds×t+b×dt ds=τbκt
We can think of the vector b b b\boldsymbol{b}b as measuring the rate of change of the osculating plane as we traverse the curve. This in turn is a measure of the rate at which the curve deviates from being a curve confined to a plane.
We now have a useful triplet to describe curves in three dimensions in terms of how the three quantities vary along the curve. Everything goes together to form 9 9 ^(9){ }^{9}9 the Frenet-Serret equations. They are written in matrix form as
(30.16) d d s ( t p b ) = ( 0 κ 0 κ 0 τ 0 τ 0 ) ( t p b ) (30.16) d d s t p b = 0 κ 0 κ 0 τ 0 τ 0 t p b {:(30.16)(d)/((d)s)([t],[p],[b])=([0,kappa,0],[-kappa,0,tau],[0,-tau,0])([t],[p],[b]):}\frac{\mathrm{d}}{\mathrm{~d} s}\left(\begin{array}{c} \boldsymbol{t} \tag{30.16}\\ \boldsymbol{p} \\ \boldsymbol{b} \end{array}\right)=\left(\begin{array}{ccc} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{array}\right)\left(\begin{array}{c} \boldsymbol{t} \\ \boldsymbol{p} \\ \boldsymbol{b} \end{array}\right)(30.16)d ds(tpb)=(0κ0κ0τ0τ0)(tpb)
These equations summarize the curvature of paths in three-dimensions and are the key results, so far, from our investigation of classical curvature. They show that we only need two parameters: κ κ kappa\kappaκ and τ τ tau\tauτ.
8 8 ^(8){ }^{8}8 In this chapter, from here on, we will denote d / d s d / d s d//ds\mathrm{d} / \mathrm{d} sd/ds by a dot.
By cyclic permutation, we have
b = t × p t = p × b (30.11) p = b × t b = t × p t = p × b (30.11) p = b × t {:[b=t xx p],[t=p xx b],[(30.11)p=b xx t]:}\begin{align*} & b=t \times p \\ & t=p \times b \\ & p=b \times t \tag{30.11} \end{align*}b=t×pt=p×b(30.11)p=b×t
9 9 ^(9){ }^{9}9 Jean Frédéric Frenet (1816-1900), Joseph Alfred Serret (1819-1885). The equations were obtained by Serret in 1851, but can be found in Frenet's 1847 thesis, an abstract of which was published in 1852 . As a result of their slightly complicated birth, they are sometimes called the Serret-Frenet equations.
10 10 ^(10){ }^{10}10 Leonhard Euler (1707-1783). "The study of Euler's works will remain the best school for the different fields of mathematics, and nothing else can remathematics, and nothing else can re-
place it." So said Johann Carl Friedrich place it." So said Johann Carl Friedrich
Gauss (1777-1855), another giant of mathematics whose work we will discuss shortly.
Fig. 30.6 A curved, two-dimensional surface is embedded in three dimensions. We measure the curvature of the point using its tangent plane. The the point using its tangent plane. The
height of the curve above the tangent plane is z z zzz.